Download Ch. 15: Electric Forces and Electric Fields

CH. 15: ELECTRIC FORCES AND ELECTRIC
FIELDS
The first evidence for electric forces was the observation that, after rubbing with wool,
silk, or hair, some objects are able to attract other objects, like a comb attracting pieces of
paper, or a balloon sticking to a wall. Three thousand years ago, a thousand years ago,
and up until just a few hundred years ago, this behavior must have seemed odd, almost
magical. We now understand that it is simply explained as the transfer of some electrons
between the two rubbed objects, resulting in a net charge on the comb or balloon which
produces an electric force on other objects.
15.1 Properties of Electric Charge
Electric charge comes in two types, positive and negative. The nuclei of atoms are
positively charged, as are protons, and cations. Electrons and anions are negatively
charged.
There are a few basic rules for charges:
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Like charges repel; unlike charges attract.
Electric charge is always conserved. Charge cannot be created or destroyed.
Electric charge is quantized. That is, there is a smallest unit of charge, and, for our
purposes, it is equal to the charge on a single electron or proton.
The last rule means that we can only have charges that equal an integral number of
electrons or protons. We can never have a fraction of an electron charge. The electron
charge is very small, and when dealing with larger quantities of charge, we usually
neglect the discreteness of charge and treat it as a continuously varying quantity -- the
same as we do with mass.
The SI unit of charge is the coulomb, abbreviated C. The charge of a single electron is 1.60219×10-19C, or put another way, -1.0C of charge contains 6.24×1018 electrons. While
much less than a mole of electrons (6.02×1023), this is nevertheless a substantial number.
The same holds true for protons if the sign of charge is made positive.
15.2 Insulators and Conductors
It is found that most materials fall into one of two classes distinguished by how freely
charge is able to move in the material. In conductors charges can move about freely.
Most metals are good conductors. On the other hand, charges in insulators are basically
fixed, not free to move. Glass, rubber, and most plastics are insulators.
While some conductors are better conductors than others, and some insulators are better
insulators than others, dividing the world into these two broad categories is a good
starting point. Differentiating between degrees of conductors and insulators is just a
refinement of thee broad categories.
A small number of materials fall into a third category between conductors and insulators.
These are known as semiconductors, the most widely known example being silicon.
15.3 Coulomb's Law
After people realized that electric charges exist (a major contribution was made by
Benjamin Franklin), experiments were undertaken to understand the force exerted by one
charge on another. The most detailed experiments were done by Charles Coulomb. He
observed that the electric force is:
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

inversely proportional to the square of the separation between the particles and
directed along the line joining them,
proportional to each of the charges,
attractive for charges of opposite sign, and repulsive for charges of the same sign.
These properties are summarized in an equation known as Coulomb's law:
F = k |q1| |q2| / r²
where k = 9×109Nm²/C² is the Coulomb constant, q1 and q2 are the two charges in
coulombs, and r is the distance between them in meters.
Notice the similarities between the electrical force and the gravitational force.
The electrical force is a force just like all those encountered in Physics 2130 (spring
force, normal force, gravitational force). As such, the electrical force obeys Newton's
third law (for every action there's an equal and opposite reaction). Therefore, for a pair of
charges, 1 and 2, F12 = -F21.
Example P15.4
An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge =
+79e). What is the electrical force acting on the alpha particle when it is 2.0×10-14m from
the gold nucleus?
We'll use Coulomb's law to calculate the electrical force. F = kq1q2/r² =
(9×109Nm²/C²)(2.0e)(79e)/ (2.0×10-14m)² = (360×1019N/c²)(1.6×10-19C)2 = 9.1×10-17N.
Example P15.8
A 2.2×10-9C charge is on the x axis at x = -1.5m and a 5.4×10-9C charge is on the x axis
at x = 2.0m. Find the net force exerted on a 3.5×10-9C charge located at the origin.
Start by making a sketch of the problem. Draw the x axis and the locations of the 3
charges. Show the forces exerted on the charge at the origin; there are 2 of them, and
since all the charges are positive, all the forces are repulsive. One force points left and the
other points right.
F1 = kq1Q/r²1 = (9.0×109Nm²/C²)(2.2×10-9C)(3.5×10-9C)/(1.5m)² = 31×10-9N = 3.1×108
N. F1 points to the right (+x direction).
F2 = kq2Q/r²2 = (9.0×109Nm²/C²)(5.4×10-9C)(3.5×10-9C)/(2.0m)² = 43×10-9N = 4.3×108
N. F2 points to the left (-x direction).
The net force is Ftot = F1 - F2 = -1.2×10-8N.
Recall from last lecture:
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


Electric charge:
o can be positive or negative
o comes in discrete quantities
o is conserved
o like signs repel, opposite signs attract.
Most materials fall into one of two categories:
o insulators -- charges are fixed
o conductors -- charges are free to move.
Coulomb's Law (for point charges or spheres): F = k |q1| |q2| / r²
The Coulomb force is directed along the line connecting the centers of the two
charges.
Example P15.4
An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge =
+79e). What is the electrical force acting on the alpha particle when it is 2.0×10-14m from
the gold nucleus?
First, sketch an alpha particle moving towards a gold nucleus. For the purposes of this
course, electrons, protons, and nuclei can be considered to be point charges. An alpha
particle is a helium nucleus.
We'll use Coulomb's law to calculate the electrical force.
F = kq1q2/r² = (9×109Nm²/C²)(2.0e)(79e)/(2.0×10-14m)² = (360×1037N/c²)(1.6×10-19C)2 =
91N. This force is repulsive since both charges (alpha particle and gold nucleus) are
positive.
The Superposition Principle
When dealing with 3 or more charges, the net force on one of them is the vector sum of
the forces due to each of the others taken individually. For example, if we have charges
called 1, 2, 3 ... the total force exerted on charge 1 is:
Ftot = F12 + F13 + ...
F12 is the force on charge 1 from charge 2. This is a vector sum; you must sum the x and
y components of the forces separately.
Example P15.8
A 2.2×10-9C charge is on the x axis at x = -1.5m and a 5.4×10-9C charge is on the x axis
at x = 2.0m. Find the net force exerted on a 3.5×10-9C charge located at the origin.
Start by making a sketch of the problem. Draw the x axis and the locations of the 3
charges. Show the forces exerted on the charge at the origin; there are 2 of them, and
since all the charges are positive, all the forces are repulsive. One force points left and the
other points right.
F1 = kq1Q/r²1 = (9.0×109Nm²/C²)(2.2×10-9C)(3.5×10-9C)/(1.5m)² = 31×10-9N = 3.1×108
N. F1 points to the right (+x direction).
F2 = kq2Q/r²2 = (9.0×109Nm²/C²)(5.4×10-9C)(3.5×10-9C)/(2.0m)² = 43×10-9N = 4.3×108
N. F2 points to the left (-x direction).
The net force is Ftot = F1 - F2 = -1.2×10-8N. F2 is subtracted because it points to the left, in
the -x direction.
15.4 The Electric Field
The electric field is something that upon first sight seems like little more than slight of
hand. We define the electric field at a point in space to be the force that would be
produced on a test charge at that location, divided by the value of the test charge:
E = |F| / |q0|
where q0 is the test charge, and F is the force on it.
So what have we gained? At first sight, very little. However, as we proceed through the
study of electricity and magnetism, the electric field takes on a reality of its own, until
joining with the magnetic field in electromagnetic waves.
Since the force on a test charge q0 from a charge q is F = |q| }q0| / r², the electric field due
to q is:
E = k |q| / r².
The electric field is a vector quantity, and, by convention, it points in the same direction
as would the force produced on a positive test charge. The electric field has units of N/C.
The electric field obeys the superposition principle exactly as do forces. Therefore, the
net electric field due to two or more charges is given by summing the electric field
produced by each charge individually. Again, you must sum the x and y components
because E is a vector quantity.
Example: P15.8 (modified)
A 2.2×10-9C charge is on the x axis at x = -1.5m and a 5.4×10-9C charge is on the x axis
at x = 2.0m. Find the electric field at the origin.
Label the charge at -1.5m number 1, and the charge at 2.0m number 2. Determine the
electric field due to each, and sum it up.
E1 = k q1 / r1² = (9.0×109 Nm²/C²)(2.2×10-9C)/(1.5m)² = 8.8 N/C. E1 points in the +x
direction.
E2 = k q2 / r2² = (9.0×109 Nm²/C²)(5.4×10-9C)/(2.0m)² = 12.2 N/C. E2 points in the -x
direction.
Etot = E1 - E2 = 8.8 N/C - 12.2 N/C = -3.4 N/C.
What is the force on a 3.5×10-9C charge at the origin?
F = qE = (3.5×10-9C)(-3.4 N/C) = -12×10-9 N. This is the same value we calculated
earlier, as it must be.
Recall from last lecture:
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
Coulomb's law: F = k q1q2 / r²
Electric field: E = k q / r²
Both the electric force and electric field are vector quantities.
The above relations are valid for point charges or spherical charges.
For a collection of charges, the total force or field is given by summing the
contributions from each charge individually -- the superposition principle.
15.5 Electric Field Lines
We can graphically represent the electric field at all points in space by drawing electric
field lines. A drawing of the electric field lines immediately shows several things.
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
The electric field vector, E, is tangent to the electric field lines at each point.
The magnitude of the electric field is proportional to the density of field lines near
a particular location.
The electric field lines of an isolated positive charge point radially outward. For an
isolated negative charge the lines point radially inward. (Draw examples.) For more
complicated arrangements of charge, the lines are normally curved, not straight.
There are some rules to follow for drawing electric field lines:
1. The lines must begin on positive charges (or at infinity), and end on negative
charges (or at infinity).
2. The number of lines drawn leaving a positive charge or entering a negative charge
is proportional to the magnitude of the charge.
3. No fields lines can cross.
(Draw the field lines for a pair of charges +2q and -q.)
15.6 Conductors in Electrostatic Equilibrium
The ability of charges to move freely in a conductor results in some special properties.
1. The electric field is zero everywhere inside the conductor.
2. Any excess charge on an isolated conductor resides entirely on its surface.
3. The electric field just outside a charged conductor is perpendicular to the
conductor's surface.
4. On an irregularly shaped conductor, the charge tends to accumulate at locations
where the radius of curvature of the surface is smallest, that is, at sharp points.
The first property is true since if the electric field is not zero inside a conductor, then
some of the sea of mobile charges inside would move until the electric field became zero.
The second property is a result of the 1/r² behavior of the electrical force.
The third property becomes evident after considering what happens if it is not true. If the
field just outside the conductor has a component parallel to the surface of the conductor,
then it will produce a force on surface charges that is parallel to the surface. These
charges are free to move along the surface of the conductor (but not off the conductor),
and so they will, until the parallel component of the force vanishes. It is important to
recall that the electric field not only acts on the surface charges, it is produced by the
surface charges. The movement of the charges changes the electric field until equilibrium
is established.
The fourth property is harder to prove precisely, but is the reason why it can be
dangerous to be on a hilltop or mountaintop during a thunderstorm. In such locations, a
standing person can become a "sharp" point, resulting in a concentration of electric field
lines at the person, and making him/her a target for a lightning strike.
I will skip sections 15.8 and 15.9.
15.10 Electric Flux
From section 10 we will be discussing electric flux. Here we will learn only how flux is
calculated, and won't make further use of it. The primary reason to discuss this now is to
learn about how flux is calculated to prepare for when we meet it again in the case of
magnetic fields.
Consider a region of uniform electric field, and a surface of area A perpendicular to the
field. We define the flux as the amount of electric field passing through this area:
=EA
where E is the value of the electric field in N/C and A is the area in m². The flux has units
of Nm²/C. If the surface is not perpendicular to the field then the flux is given by:
 = E A cos
where  is the angle between the perpendicular to the surface and the electric field.
Example P15.39
A 40cm diameter loop is rotated in a uniform electric field until the position of maximum
electric flux is found. The flux in this position is measured to be 5.2×105Nm²/C.
Calculate the electric field strength in this region.
Begin by drawing a sketch of the uniform electric field, and a circular loop. The loop
forms the boundary of a circular surface of area A = π(0.20m)² = 0.126m². Using the
definition for flux, we have  = E A cos. Since E and A are fixed, the flux is maximum
when cos = 1. This occurs when the loop is perpendicular to the field. Now we can solve
for the electric field.
E = /A = (5.2×105Nm²/C)/0.126m² = 41×105 = 4.1×106.
Example: Flux through half a sphere
Calculate the flux through half a spherical shell of radius 10cm due to a charge of -4×109
C located at the center of the shell.
The electric field from a point charge is constant on a spherical shell centered on the
charge. Also, the electric field points radially outward; it is perpendicular to the surface
of the spherical shell at every point where it penetrates the shell. Therefore, even though
the surface is not flat, we can still calculate the flux as the product of the field times the
area. The area of half a spherical shell is A = 2πr². The electric field is E = k q / r². The
flux is
 = E A = (k q / r²)(2πr²) = 2πkq = 2π(9×109Nm²/C²)(-4×10-9C) = -226Nm²/C.
Recall from last lecture:
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

Electric field: E = k q / r²
Rules for drawing electric field lines.
Properties of conductors in electrostatic equilibrium.
CH. 16: ELECTRICAL ENERGY AND
CAPACITANCE
Recall that the concept of conservative forces, work, and potential energy is useful in
solving mechanics problems. It is often possible to solve problems by applying
conservation of energy, and arrive at the solution faster and easier than by solving for the
motion. The same idea holds with the electrostatic force. In this chapter we will learn
about electrical energy, and apply it to a common electrical device called a capacitor.
16.1 Potential Difference and Electrical Potential
Recall that the gravitational force, Fg = G m M / r², is a conservative force because the
work done to move a particle from point A to point B in a gravitational field depends
only on the locations A and B, but not on the path taken from A to B. Because the
electrostatic force has the same form as the gravitational force, the electrostatic force is
also conservative. For a conservative force, we can define a potential energy function.
Recall that we defined two potential energy functions for the gravitational force: one for
the case of a uniform gravitational field, like near the surface of the earth, PE = mgh;
and one for the case of a spherical mass, like a planet, PE = -G M m / r. (Potential
energy and work are measured in joules (J), as is energy.) We will do likewise for the
electrical force.
Potential energy in a uniform electrical field
First consider the case of a uniform electrical field, E. Imagine that we move a charge q
from point A to B, a distance d parallel to the field. (Provide a drawing.) The force on the
charge is F = qE, and is directed parallel to the field. The work done is W = F cos s
(Equation 5.1), where  is the angle between the force and the direction of displacement.
The charge is moved a distance d parallel to the force, so we have:
W = Fd = qEd
The change in potential energy is the negative of the work done:
PE = -W = -qEd
This is the potential energy function for a uniform electric field.
In electric circuits we will frequently be dealing with varying amounts of charge. For this
reason, it is even more useful to define something we call the electric potential:
V = VB - VA = PE / q
Electric potential is a scalar, a number (like potential energy and work), and it is
measured in units of J/C or the more familiar volt (V). 1V = 1J/C. If a charge q moves
through an electric potential, V, the change in potential energy is PE = qV.
A couple of words about units.
Since V = PE / q = -qEd / q = -Ed (in the case of uniform field). This can also be
written as E = -V/d. Therefore we can write the following relation between units: 1N/C =
1V/m. Electric fields are often given the units of V/m.
Example P16.2
A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A
+12µC charge moves from the origin to the point (x,y) = (20cm, 50cm). (a) What was the
change in the potential energy of this charge? (b) Through what potential difference did
the charge move?
Begin by drawing a picture of the situation, including the direction of the electric field,
and the start and end point of the motion.
(a) The change potential energy is given by the charge times the field times the distance
moved parallel to the field. Although the charge moves 50cm in the y direction, the y
direction is perpendicular to the field. Only the 20cm moved parallel to the field in the x
direction matters for determining the change of potential energy. PE = -qEd = (+12µC)(250 V/m)(0.20 m) = -6.0×10-4 CV = -6.0×10-4 J.
(b) The potential difference IS the difference of electric potential, V. V = PE / q = 6.0×10-4 J / 12µC = -50 V.
Example P16.10
(a) Through what potential difference would an electron need to accelerate to achieve a
speed of 60% of the speed of light, starting from rest? (The speed of light is 3.00×108
m/s.)
(a) The final speed of the electron is vf = 0.6(3.00×108 m/s) = 1.80×108 m/s. At this
speed, the energy (non-relativistic) is the kinetic energy,
E = ½mvf² = 0.5(9.11×10-31 kg)(1.80×108 m/s)² = 1.48×10-14 J.
This energy must equal the change in potential energy from moving through a potential
difference, V, E = PE = qV. Therefore:
V = E/q = (1.48×10-14 J)/(-1.6×10-19 C) = -9.25×104 V.
16.2 Electric Potential and Potential Energy Due to Point Charges
The electric potential a distance r from a point charge q is given by:
V=kq/r
where zero electric potential is at infinity. (Recall that we are free to choose the zero of
potential energy.) When dealing with two or more charges, the electric potential at a point
is given by the sum of the contributions from each charge individually. This is again a
consequence of the superposition principle. Force and electric field are vectors, the
potential is a scalar, that is just a single number without direction. When summing the
contributions to the potential, do not take x and y components!
The potential energy required to bring a charge q2 from infinity to a distance r from
charge q1 is given by
PE = q2V1 = k q1 q2 / r
Example P16.14
Three charges are situated at corners of a rectangle as in Figure P16.13. How much
energy would be expended in moving the 8.0µC charge to infinity?
(Drawing) We are dealing with point charges, so we will take the zero of potential and
potential energy at infinity. Let's solve the problem using W = PE = q(VB - VA), where
q is the 8.0µC charge, A is the upper left corner of the rectangle, and B is infinity. (Note
that it doesn't matter which direction we move away to, as long as it is infinitely far away
from the charges.)
As noted above, VB = 0. Let's determine VA. It has two contributions, one from the 2.0µC
charge, V2, and one from the 4.0µC charge, V4. Recall that the prefix µ means 10-6.
V2 = k q2 / r2 = (9×109)(2.0µC)/(0.030 m) = 6.0×105 V.
V4 = k q4 / r4 = (9×109)(4.0µC)/sqrt{(0.030 m)² + (0.060 m)²} = 5.4×105 V.
Summing these 2 contributions gives a total potential:
VA = 11.4×105 V.
The energy expended is:
W = (8.0µC)(0 V - 11.4×105 V) = -91×10-1 CV = -9.1 J.
Recall from last lecture:
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

Potential Energy in a Uniform E-field: PE = -qEd
Electric Potential (Voltage): V = PE / q
Electric Potential due to a point charge: V = k q / r
16.3 Potentials and Charged Conductors
The electric potential is the same everywhere inside a charged conductor. This is because
the electric field is zero inside the conductor (see Ch. 15), and no work is done to move a
charge through zero electric field. If no work is done, then the change in electric potential
is zero, so it must be constant.
This result continues to hold at the surface of the conductor. At the surface of the
conductor, the electric field is perpendicular to the surface. Therefore, if a charge moves
along the surface, it moves perpendicular to the electric field, and perpendicular to the
electric force. No work is done on an object that moves perpendicular to a force. So again
we see that the electric potential must be constant inside and at the surface of a
conductor.
Note that the electric field is zero inside a conductor, but the electric potential need not be
zero, just constant.
The Electron Volt
A unit of energy common in atomic, nuclear, and particle physics is the electron volt,
abbreviated eV. The electron volt is the energy released (or supplied) when an electron
moves through a potential difference of 1V (-1V). Because one electron has a charge of 1.6×10-19C, and 1V = 1J/C, the eV is:
1eV = 1.6×10-19C V = 1.6×10-19J.
16.4 Equipotential Surfaces
Imagine some fixed charges, and the electric field associated with them. Now imagine
calculating the electric potential at lots of points around the charges. If we are clever, we
can find locations around the charges where the electric potential is the same, and with a
little thought, you should realize that these points form lines (or surfaces if we move to a
3 dimensional picture). These lines of equal electric potential are called equipotential
lines, and they are another useful way to help us visualize the electric field. (Make
sketches like those in Fig. 16.8) Notice that where the equipotential lines are always
perpendicular to the electric field lines.
16.5 Applications
The text reviews two interesting devices that make use of the action of electric fields on
charges. The first is the electrostatic precipitator. An electrostatic precipitator, commonly
part of "scrubbers", is a device for removing particulate matter from combustion gases. It
consists of a chamber with a high electric field inside. The electric field serves both to
ionize the particles, and to draw them to the walls of the container where they precipitate
from the gas. The second device is the photocopier/laser printer. A photocopier uses a
drum that can be electrically charged, and the charge can be fixed in a pattern that mimics
what is to be copied. Toner particles with the opposite charge adhere to the charged
regions of the drum, and then this is transferred to paper.
16.6 The Definition of Capacitance
Capacitors are devices commonly used in electric and electronic circuits. In fact you may
have seen something in the news last spring about a shortage of capacitors for cell phones
-- a typical cell phone needs several dozen capacitors.
A capacitor has a rather simple construction: it looks like two identical conducting plates
separated by a small gap, each plate can be connected to some device for generating a
charge. We will consider only cases where the plates have equal amounts of oppositely
signed charge. So, our standard picture of a capacitor is of two plates, each with area A,
separated by a small gap, d, one with charge +Q and the other with charge -Q.
Capacitance, C, is defined as the charge on one conductor divided by the potential
difference between them:
C = Q/V
The units of capacitance are, from the equation, charge over voltage. This ratio is called
the farad, 1 F = 1 C/V. Capacitors typically come in values of microfarads (1 µF = 10-6F)
or picofarads (1 pF = 10-12F).
Note: We are running short on letters. The letter C stands for coulomb (charge) when
used as a unit, and stands for capacitance when used as a symbol. You must determine
which it is from the context. If C appears in an equation or other expression, it probably
stands for capacitance. If C appears following a value, as in 12µC, it probably stands for
coulomb.
Example 16.4
A 3.0µF capacitor is connected to a 12V battery. What is the magnitude of the charge on
each plate of the capacitor?
Use the definition of capacitance and solve for Q:
Q = CV = (3.0µF)(12V) = 36µC
Note that a 12V battery has a potential difference of 12V between its terminals, and when
each terminal is connected to a capacitor plate via a conducting wire, the plates will have
a potential difference of 12V.
16.7 The Parallel-Plate Capacitor
The parallel-plate capacitor was described above. A nice feature of this design (and the
reason it is used as our standard capacitor) is that the capacitance can be determined by
the construction. The capacitance depends on the area of the plates, A,the size of the gap,
d, and the insulator between the plates. For now we will assume that the insulator is air,
then:
C = 0 A / d
where 0 is a constant called the permittivity of free space. It is related to the Coulomb
constant, k, by k = 1/(40). From this relation we can determine the value of 0 = 1/(4πk)
= 1/(4π(9×109Nm²/C²)) = 8.85×10-12 C²/Nm²
From the equation for a parallel-plate capacitor we see that the capacitance will increase
if the area of the plates is increased, or the gap between them is decreased. The
capacitance will decrease if the area is decreased or the gap is increased. For example, if
the spacing between the plates of a capacitor is reduced to half its original value, the
capacitance will double. (This effect is used in computer keyboards and many types of
sensors.)
Example: P16.22
Consider the Earth and a cloud layer 800m above the Earth to be the plates of a parallelplate capacitor. (a) If the cloud layer has an area of 1.0km² = 1.0×106 m², what is the
capacitance? (b) If an electric field strength greater than 3.0×106 N/C causes the air to
break down and conduct charge (lightning), what is the maximum charge the cloud can
hold?
(a) Assume the effective conducting are on the Earth matches the area of the cloud.
C = 0A/d = (8.85×10-12 C²/Nm²)(1.0×106 m²)/(800m) = 1.1×10-8 F.
A parallel plate capacitor has a uniform electric field between its plates, and zero field
outside. The potential difference when moving a distance d through a uniform E-field is
V = Ed. In this case, the distance is gap between the plates of the capacitor, or the
distance from the Earth to the clouds. The charge needed to produce that potential
difference is
Q = CV = CEd = (1.1×10-8 F)(3.0×106 N/C)(800m) = 26 C.
Recall from last lecture:




Conductors: Electric field is zero inside, and perpendicular to the surface; electric
potential is constant inside and on surface.
Equipotential Surfaces: Surfaces across which the potential is a constant value;
perpendicular to electric field lines.
Capacitance: C = Q/V
Parallel-Plate Capacitor: C = 0 A / d
Symbols for Circuit Elements
We will now start drawing electric circuits. These drawings are called schematics
because they are a representation of the circuit in an electrical sense, not a physical sense.
Electric circuits are composed of circuit elements (batteries, capacitors, resistors, etc.)
connected together by conductors represented by lines. Symbols representing capacitors,
batteries, and resistors are shown in section 16.7 of the text.
16.8 Combinations of Capacitors
We want to consider how to treat combinations of capacitors. We begin by considering
cases with two capacitors combined, and from these we can build up any arbitrarily
complex combination. There are only two ways to combine two capacitors, we call them
parallel combination and series combination. We will derive an expression for a single
capacitance that is equivalent to the combination of capacitors, and in the process we will
apply what we have learned about charges, potential difference (voltage), and conductors.
Parallel Combination
(Draw a parallel combination with capacitors C1 and C2 connected to a battery.) In a
parallel combination, the capacitors are usually drawn side by side. If we imagine them as
parallel-plate capacitors with the same gap, snuggling them right up next to each other,
the combination seems to become a single capacitor with an area equal to the sum of the
areas. Then from the equation for capacitance of a parallel-plate capacitor, we have
Ceq = 0Aeq/d = 0(A1+A2)/d = 0A1/d + 0A2/d = C1 + C2
Thus, Ceq = C1 + C2.
Let's do this again, but this time thinking in terms of charges and voltage because it will
probe useful for the series combination. When connected in parallel, the voltage across
each capacitor is the same: V1 = V2 = V. The charge on each capacitor is
Q1 = C1V and Q2 = C2V
So the total charge is
Q = Q1 + Q2 = C1V + C2V = ( C1 + C2 )V
The equivalent capacitance is defined to be
Ceq = Q/V = C1 + C2
the same result as above.
It is pretty clear how to extend this result to more than two capacitors in parallel. In that
case
Ceq = C1 + C2 + C3 + ...
Note that the equivalent capacitance of a parallel combination is always greater than any
of the individual capacitances in the combination.
Series Combination
(Draw a series combination.) In a series combination, the capacitors are connected headto-tail. We want to replace the pair by a single equivalent capacitor (draw equivalent
capacitor circuit). To do this, we must understand how the charge is distributed on the
plates.
Consider the inner pair of plates, one from each capacitor, connected by a conductor.
These three objects are electrically isolated from the remainder of the circuit -- they form
a single isolated conductor. Since the net charge on the capacitors is zero before the
battery is connected, the net charge on the inner pair of plates must also be zero. After the
battery is connected, the plates of the capacitors will hold some charge, but the inner pair
of plates will still have zero net charge. Therefore, the charges on the inner pair of plates
is equal and opposite, and we see that both capacitors will hold the same charge.
We don't add these charges together, as in the parallel case. The quantity that adds is the
voltage across each capacitor
V = V1 + V2
The voltage across the capacitors is related to their charge
V1 = Q/C1 and V2 = Q/C2
The definition of the equivalent capacitance is Ceq = Q/V, or V = Q/Ceq. Therefore
Q/Ceq = Q/C1 + Q/C2
We can divide out the common factor of Q to arrive at
1/Ceq = 1/C1 + 1/C2
It is not too difficult to see that for three or more capacitors in series, the equivalent
capacitance is given by:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...
Example P16.31
Consider the combination of capacitors in Figure P16.31. (a) What is the equivalent
capacitance of the group? (b) Determine the charge on each capacitor.
(a) This circuit consists of a three parallel branches, the first two branches are single
capacitors, and the last branch is a series combination of two capacitors. Begin by finding
the equivalent capacitance of those two, then the equivalent capacitance of the three
parallel combination can be found.
1/C3eq = 1/24µF + 1/8.0µF = 1/6.0µF
Therefore, C3eq = 6.0µF. Remember to take the inverse to get the equivalent capacitance
of a series combination. Now, the parallel combination gives:
Ceq = 4.0µF + 2.0µF + 6.0µF = 12.0µF
(b) The charge on each depends on the voltage across the capacitor. The voltage across
the 4 and 2µF capacitors is the full 36V of the battery.
Q4 = (4.0µF)(36V) = 144µC
Q2 = (2.0µF)(36V) = 72µC
The charge on the 2 series capacitors is the same, and is equal to the charge that would
exist on their equivalent value.
Q24 = Q8 = Q3eq = (6.0µF)(36V) = 216µC
16.9 Energy Stored in a Charged Capacitor
To determine how much energy is stored on a capacitor, we imagine starting with a
neutral capacitor, and adding a little charge, then a little more, then a little more, each
time adding up how much energy is required. The first bit of charge takes no energy, but
the next little bit must be added to a capacitor that's now has a potential difference, V =
CQ, and requires an energy VQ. For the next bit of charge, the potential is a bit higher,
and a little more energy is needed. When all the energy is summed up, the total energy
required to fully charge a capacitor to a potential difference V is:
Estored = ½QV = ½CV² = Q² / 2 C
You can make use of whichever expression is most convenient. This result holds for any
capacitor, not only parallel-plate capacitors.
Example P16.40
Consider the parallel-plate capacitor formed by the Earth and a cloud layer as described
in problem 22. Assume this capacitor will discharge (lightning) when the electric field
strength between the plates reaches 3.0×106N/C. What is the energy released if the
capacitor discharges completely during a lightning strike?
From last lecture, we found the capacitance of the cloud 800m above the Earth to be C =
1.1×10-8 F. When the capacitor discharges completely, the energy stored is zero.
Therefore, the energy released is equal to the energy stored before it discharges. Since we
previously determined that the charge stored is Q = 26 C, let's use Estored = Q² / 2 C to
determine the energy stored:
Estored = (26 C)² / 2(1.1×10-8 F) =
Note that we could also have calculated this result using Estored = ½CV², and V = Ed.
Estored = ½C(Ed)² = ½(1.1×10-8 F)(3.0×106N/C)²(800m)² =
16.10 Capacitors with Dielectrics
A dielectric is an insulating material. When such an insulating material is placed between
the plates of a capacitor, the atoms and molecules tend to polarize, and the capacitance
increases. The entire effect is handled by a number called the dielectric constant for the
material used, . A list of dielectric constants is given in Table 16.1 in the text. When a
dielectric is used, the capacitance is changed from what it would be with no dielectric
material by:
C = C0
where C0 is the capacitance with no dielectric, and C is the capacitance with the dielectric
of constant .
Recall:



Capacitances in parallel -- add.
Capacitances in series -- inverses add
Electric circuits and symbols.
CH. 17: CURRENT AND RESISTANCE
We are now moving from phenomena with fixed charges to phenomena with moving
charges. Electric and electronic devices operate with moving charges. The material in this
chapter is basic knowledge we will need and use in the study of electric circuits in the
next chapter.
17.1 Electric Current
Electrons are constantly moving about inside a conductor, jiggling in all directions, with
zero net motion. Electric current flows when some force acts on the electrons resulting in
net motion. Note the similarity with the kinetic theory of gases.
Imagine a wire, and a surface that intersects the wire. We define the current through the
wire to be the net flow of charge across the surface per second:
I = Q / t
where Q is the net charge flowing across the surface in time t. The charge is counted
such that:


a positive charge crossing in the positive direction, or
a negative charge crossing in the negative direction
count as positive Q, while:


a positive charge crossing in the negative direction, or
a negative charge crossing in the positive direction
count as negative Q. The unit of current is the ampere, written A, with
1 A = 1 C/s.
In a circuit, we will want to know the direction of current. Although we now know that,
with few exceptions, current is due to the flow of electrons, we still define the direction
of current as the direction of flow of positive charge. This is opposite the direction of
electron flow.
Example: P17.4
In a particular television picture tube, the measured beam current is 60.0µA. How many
electrons strike the screen every second?
The picture tube of a television or computer monitor operates by shooting a beam of
electrons from an electron source (usually a hot filament) at a phosphor coated screen.
The phosphor is excited by the electrons and emits light when the atom relaxes back to its
ground state. The intensity of the light is proportional to the intensity of the electron
beam.
A beam current of 60.0µA means that 60µC of charge strike the screen per second. This
is equivalent to (60.0×10-6C)/(1.6×10-19C/e) = 37.5×1013 electrons = 3.75×1014 electrons.
17.2 Current and Drift Speed
It is instructive to relate macroscopic current to the microscopic properties of conductors.
Typically, most of the electrons in a material are strongly bound to atoms -- these are the
inner shell electrons -- and only a fraction are free to move, the outermost electrons. The
free electrons we call mobile charges.
Consider a wire with cross section A. Let n be the density of mobile charges (the number
per unit volume). The number of mobile charges in a length of wire x is the density
times volume or nAx. If each mobile charge has charge q (or the average is q), then the
total amount of mobile charge in length x is
Q = (nAx)q
if the charges move with average speed vd, then in time t they move a distance x =
vdt. Then, the amount of charge that moves across a cross-sectional surface of the wire
in time t is:
Q = (nAvdx)q.
The current due to this movement of charge equals:
I = Q / t = nqvdA
This relation relates the current (a macroscopic quantity) to the density of mobile charges,
their charge, their average drift speed, and the area of the wire.
Example: P17.7
A 200km long high-voltage transmission line 2.0cm in diameter carries a steady current
of 1000A. If the conductor is copper with a free charge density of 8.5×1028 electrons per
cubic meter, how long (in years) does it take one electron to travel the full length of the
cable?
Begin by finding the drift velocity of the electrons.
vd = I / nqA = (1000A)/(8.5×1028e/m³)(1.6×10-19C/e)π(0.020m)² = 5.85×10-5m/s
Now determine how long it will take to travel 200km at this speed. Use 1year = 3.1×107s.
t = l / vd = (2.0×105m) / (5.85×10-5m/s) = 0.34×1010s/ (3.1×107s/year) = 110 years.
17.3 Resistance and Ohm's Law
An electric field is required to cause a net flow of electrons. The electrons are accelerated
parallel to the field until they hit an atom in the material, at which point they transfer
most of their energy to the atom, and bounce off in some other direction. The atoms tend
to slow down the motion of the electrons, acting like a drag force. The macroscopic effect
of this drag force is termed resistance.
We define the resistance, R, as the ratio of the applied voltage to the resulting current:
R = V / I
For many materials, this relationship works well over a wide range of voltage, current,
temperature, humidity, etc. (Later we will see how to make corrections for the change in
resistance with temperature.) The unit of resistance is the ohm, written ,
1 = 1 V/A.
It is rather a coincidence that many materials exhibit a constant resistance over a wide
range of conditions. Resistances from values of µ = 10-6 to G = 109 are available.
The fact that voltage and current are related by a constant is known as Ohm's law,
normally expressed as:
V = IR.
Example 17.3 The Resistance of a Steam Iron
All electric devices are required to have identifying plates that specify their electrical
characteristics. The plate on a certain steam iron states that the iron carries a current of
6.4A when connected to a 120V source. What is the resistance of the steam iron?
Use Ohm's law:
R = V / I = (120V)/(6.4A) = 19
Recall:




current: I = Q / t, measured in amperes, 1 A = 1 C/s
drift speed: I = nqvdA
Resistance, R, measured in ohms, 1  = 1 V/A
Ohm's law: V = IR
17.4 Resistivity
Resistance is due to a "drag force" on the electrons moving through a material. This drag
force arises from the electrons bumping into atoms as they move through the material.
This force is due to the atoms in the material, not to the physical dimensions of the
material (length and cross-sectional area).
The resistance of an object is due to its dimensions. If we make the object twice as long,
the resistance will double; if we make the cross-section twice as big, the resistance will
be half. If we factor out the physical dimensions, then we are left with something called
the resistivity. The resistivity, , is a property of the material. Resistance and resistivity
are related by:
R =  l / A.
The SI units of resistivity are m, and the values for a number of materials are listed in
table 17.1 of the text.
Example: P17.12
Calculate the diameter of a 2.0cm length of tungsten filament in a small light bulb if its
resistance is 0.050.
Use R =  l/A, where A = r², and the diameter, d = 2r. Solving for A, A = l/R. From
Table 17.1 we see that tungsten has a resistivity,  = 5.6×10-8m. Thus,
A = (0.020m)(5.6×10-8m)/(0.050) = 2.24×10-8m²
The diameter is therefore: br > d = Sqrt{4A/} = Sqrt{4(2.24×10-8m²)/} = 1.7×10-4m =
0.17mm.
17.5 Temperature Variation of Resistance
Earlier I said that, for many materials, resistance is roughly constant over a wide range of
voltage, current, temperature, humidity, etc. While it is roughly true, for most materials it
is not exactly true. Normally the resistance and resistivity vary somewhat depending on
conditions, but especially with temperature.
Because the variation is small, it is approximately linear. The change of resistivity with
temperature is approximately:
 = 0[1 + (T - T0)]
where  is the resistivity at temperature T (in Celsius degrees), 0 is the resistivity at
temperature T0, and  is the temperature coefficient of resistivity. Values of  for a
number of materials is listed in Table 17.1 of the text.  is normally a positive number,
such that the resistivity tends to increase with temperature.
The relationship between resistivity and resistance allows us to write a similar
approximation for resistance:
R = R0 [1 + (T - T0)]
Example: P17.26
A wire 3.00m long and 0.450mm² in cross-sectional area has a resistance of 41.0 at
20.0°C. If its resistance increases to 41.4 at 29.0°C, what is the temperature coefficient
of resistivity?
This problem is an example where more information is given than is needed to solve the
problem. Because the length and cross-sectional area are given, you may be tempted to
start by calculating the resistivity in each case, but this is not necessary because the
temperature coefficient for change of resistance is the same as for resistivity. Simply use
the relation for the change of resistance and solve the problem directly.
I'll use R = R0[1 + (T - T0)], and solve for .
 = (R/R0 - 1)/(T - T0) = ((41.4/41.0) -1)/(29.0 - 20.0°C) = 0.00108/°C = 1.08×10-3/°C.
Example: P17.18
(a) A 34.5m length of copper wire at 20°C has a radius of 0.25mm. If a potential
difference of 9.0V is applied across the length of the wire, determine the current in the
wire. (b) If the wire is heated to 30.0°C and the 9.0V potential difference is maintained,
what is the resulting current in the wire?
(a) The description of the wire is enough information for us to determine the resistance,
and then we can use Ohm's law to determine the current for V = 9.0V. For copper,  =
1.7×10-8m, from Table 17.1. Determine the resistance with
R = l/A = (1.7×10-8m)(34.5m)/(2.5×10-4m)² = 3.0.
Use Ohm's law to find the current:
I = V/R = 9.0V / 3.0 = 3.0A.
(b) The resistance increases when the wire is heated. From Table 17.1, the temperature
coefficient for copper is  = 3.9×10-3/°C. Let R0 = 3.0 at T0 = 20.0°C, as found in part
(a), then at T = 30.0°C,
R = R0[1+(T - T0)] = (3.0)[1 + (3.9×10-3/°C)(30.0 - 20.0°C)] = 3.1.
The current decreases to
I = V/R = (9.0V)/(3.1) = 2.9A.
17.7 Electrical Energy and Power
Note: In the second paragraph, the text should read "As the charge moves from A to B
through the battery, its electrical potential energy increases by the amount QV and ..."
As charge moves around an electrical circuit, energy is transferred from a battery (source
of voltage) to resistors. Since the current is moving continuously (it is not a one time
occurrence), the rate of energy transfer is a more useful quantity. The rate of energy
transfer in a circuit is called the power, P, and is given by
P = IV = I²R = (V)²/R
where the last two relations are obtained from the first by using Ohm's law. Recall that
power is measured in watts, written W, and 1W = 1J/s.
Recall from earlier:


Ohm's law: V = I R
Power: P = I V = I² R = V²/R
In the above, I have used V in place of V.
17.7 Electrical Energy and Power (cont'd)
You may have seen the quantity kilowatt-hour written on an electric meter or bill. The
kilowatt-hour is a measure of energy -- the electric company bills us for the amount of
energy we use, not the amount of power. A kilowatt-hour (kWh) is the amount of energy
consumed by a 1kW = 1000W device operating for 1 hour = 3600s, thus 1kWh =
(1000W)(3600s) = 3.6×106J.
Example: P17.38
How much does it cost to watch a complete 21 hour long World Series on a 90.0W
television set? Assume that electricity costs $0.0700/kWh.
This problem is easiest to solve by determining how much energy is used to operate the
television in kilowatt-hours, not in joules.
Energy = (90.0W)(21h) = 1890Wh = 1.89kWh.
Cost = Energy × Rate = (1.89kWh)($0.0700/kWh) = $0.13 = 13¢
CH. 18: DIRECT CURRENT CIRCUITS
In this chapter we will build on the basic concepts of current and resistance defined in
chapter 17. We will start looking at complete electric circuits, learn how to handle
combinations of resistors, and look at circuits containing resistors and capacitors.
18.1 Sources of emf
"Sources of emf" is a phrase that simply means source of voltage. Batteries and electric
generators are common sources of voltage.
Although we think of batteries as "perfect" sources of voltage, they aren't. A more
accurate picture of a real battery is a perfect voltage source with a small resistance in
series. This resistance corresponds to an effective resistance inside the battery, not to a
separate device. For purposes of this class, we will treat batteries and generators as ideal
sources of voltage, unless specifically stated to the contrary.
18.2 Resistors in Series
We want to be able to treat arbitrary combinations of resistors. As we did with capacitors,
we will begin by considering how to handle a pair of resistors, and from this we can build
up any arbitrary combination. Once again, there are only two ways to connect two
resistors, in series, or in parallel.
We begin by treating the series combination. (Schematic of a battery with two resistors in
series, R1 and R2.) We want to determine the value of a single resistor that is equivalent
to the two individual resistors, that is, the total voltage drop and current for the single
resistor must equal the total voltage drop and current for the pair of resistors. (Schematic
of a battery with one equivalent resistor.)
The voltage from the battery must be shared between the resistors; the voltage drop
across R1 plus the voltage drop across R2 will equal the battery voltage, V:
V = V1 + V2
The current through each resistor must be equal, I1 = I2. This is due to conservation of
charge -- if the currents are not equal, then the amount of charge arriving at R2, I1t, will
not equal the amount of charge leaving R2, I2t. The result is that R2 will have a net
overall charge (and presumably R1 will have an equal but opposite charge). But the
components of a circuit have no net charge, so this simply cannot occur.
According to Ohm's law, the voltage drops across the resistors is:
V1 = IR1, and V2 = IR2.
Using these relations, the voltage across the battery is:
V = IR1 + IR2 = I(R1 + R2)
We want an equivalent resistance that gives the same current for the given voltage: V =
IReq, so we see
Req = R1 + R2 (series resistors)
We can generalize this result to the case of more than two resistors in series.
Req = R1 + R2 + R3 + ... (series resistors)
Real versus ideal batteries
Consider a real battery connected to a resistance R. Let's model the battery as an ideal
voltage source, E, with a small series resistor, r. (Here we use E for voltage to emphasize
that this is the ideal voltage source inside a real battery.) The current in this circuit is
given by
I = E/(R + r)
The voltage across the battery terminals equals the voltage that appears across the
resistor, R
V = IR = E - Ir
where I get the second form by considering the battery plus its internal resistance. The
effective battery voltage is less than the internal voltage, and decreases as the current it
supplies increases. Batteries give the largest voltage when they are supplying small
currents, and if the terminals are shorted, the current from the battery is limited.
This process of modeling real devices by combinations of ideal circuit elements is
common in electronics.
Example: P18.2
A 4.0 resistor, and 8.0 resistor, and a 12 resistor are connected in series with a 24V
battery. What are (a) the equivalent resistance and (b) the current in each resistor?
(a) Since the resistors are in series, the equivalent resistance is
Req = (4.0 + 8.0 + 12.0) = 24.0.
They are in series, so the same current passes through each, and it equals I = V/Req =
24V/24 = 1A.
18.3 Resistors in Parallel
Now consider what happens when two resistors are in parallel. (Draw schematic of a
circuit with a battery and two resistors in parallel, also draw circuit with battery and
equivalent resistance.) In this case, the voltage across the resistors is the same and in this
case equals the battery voltage, V. The current from the battery is shared between the
resistors, I1 = V/R1 through resistor R1, and I2 = V/R2 through R2. The total current from
the battery equals the sum of the resistor currents:
I = I1 + I2 = V/R1 + V/R2
The same current must flow through the equivalent resistance, I = V/Req. Equating the
currents give V/Req = V/R1 + V/R2, or cancelling the common factor V:
1/Req = 1/R1 + 1/R2 (parallel resistors)
We can extend this to the case that more than two resistors are in parallel:
1/Req = 1/R1 + 1/R2 + 1/R3 + ... (parallel resistors)
Example: P18.4
The resistors in P18.2 are connected in parallel across a 24V battery. Find (a) the
equivalent resistance and (b) the current in each resistor.
(a) Since the resistances are in parallel, the inverse of the equivalent resistance is the sum
of the inverses:
1/Req = 1/4.0 + 1/8.0 + 1/12.0 = (6 + 3 + 2)/24 = 11/24
Req = (24/11) = 2.2
(b) Resistances in parallel have the same voltage across them, but different currents. We
have to solve for each current individually:
I1 = V/R1 = 24V/4.0 = 6.0A
I2 = 24V/8.0 = 3.0A
I3 = 24V/12.0 = 2.0A
Recall:





"emf" = voltage
Resistors in series: Req = R1 + R2 + ...
Resistors in parallel: 1/Req = 1/R1 + 1/R2 + ...
For resistors in series, the equivalent resistance is greater than any of the
individual resistances.
For resistors in parallel, the equivalent resistance is less than any of the individual
resistances.
18.3 Resistors in Parallel (cont'd)
Here's a much more difficult example with a circuit containing resistors in series and
parallel.
Example: P18.13
The resistance between terminals a and b in Figure P18.13 is 75. If the resistors labeled
R have the same value, determine R.
(Draw diagram.) The resistors R are part of the resistor combination, so if we write out
the expression for the equivalent resistance, then we can relate it to R and solve for R.
This circuit is a combination of resistors in series and parallel, proceed step by step to
find the equivalent resistance. We want to find the "deepest nesting" of resistors, and
beginning there, work our way out to the whole circuit.
Begin by combining the series combination of R and the 5.0 resistor, Req1 = R + 5.0.
Next we can find the equivalent resistor for the parallel combination of the 120, 40, and
Req1 resistors:
1/Req2 = 1/120 + 1/40 + 1/(R+5) = (1+3)/120 + 1/(R+5) = 1/30 + 1/(R+5) = (R + 5 +
30)/(30(R+5)) = (R+35)/(30(R+5))
Req2 = (30R+150)/(R+35)
Finally, we sum that equivalent resistance with the second resistor, R, in series with it
(the one next to terminal a), and set the result equal to 75:
Req = 75 = R + Req2 = R + (30R+150)/(R+35) = (R² + 35R + 30R + 150)/(R + 35) = (R²
+ 65R + 150)/(R + 35)
Therefore, 75 = (R² + 65R + 150)/(R + 35). To find R, multiply both sides by (R + 35):
75(R + 35) = 75R +2625 = R² + 65R + 150
R² - 10R - 2475 = 0
This quadratic equation has two possible solutions:
R = [10 ± Sqrt{10² + 4(2475)}]/2 = [10 ± Sqrt{10000}]/2 = [10 ± 100]/2 = 5 ± 50 = 55
or -45
Resistances are positive, therefore the only logical answer is R = 55.
18.4 Kirchhoff's Rules and Complex DC Circuits
With the results for combining pairs of resistors we can analyze nearly any circuit made
from combinations of resistors and one voltage source. (The exceptions are circuits like
the "current balance".) To analyze circuits that contain more than one voltage source, or
both resistors and capacitors, we need Kirchhoff's rules. There are two rules:
1. The sum of the currents entering any junction must equal the sum of the currents
leaving.
2. The sum of the potential differences (voltages gains and drops) across all the
elements around any closed loop of the circuit must be zero.
A junction is any point in the circuit where three (or more) lines meet. (Drawing of a
simple circuit with a junction.) The conductor must remain neutral (uncharged). Where
three conductors meet to form a junction this must still hold, and that is possible only if
the amount of charge entering the junction is balanced by an equal amount leaving the
junction. Since current is just the charge moving past a location on the wire per unit time,
by dividing the amount of charge entering or leaving the junction by the time interval of
measurement, we get Kirchhoff's first rule. The first rule is equivalent to conservation of
charge, and is known as the junction rule.
A closed loop of the circuit is any path through the circuit that begins and ends at the
same point. (Indicate closed loops on the circuit drawing.) A charge moving around the
loop will gain and lose energy as it passes through circuit elements, but it must return to
its starting point with zero net change of energy. The second rule is equivalent to
conservation of energy, and is known as the loop rule.
There is a well defined procedure for applying Kirchhoff's rules to a circuit.
1. Assign names and directions to the currents in each branch of the circuit. Don't
worry if you happen to choose the direction opposite the correct choice, you will
simply get a negative value for the current. The magnitude will be correct, and the
negative sign means that the actual direction is opposite to your initial
assignment.
2. Choose the loops, and the direction to traverse each loop. Record the voltage gain
or drop for each circuit element traversed, according to the following rules:
1. If a resistor is traversed in the direction of the current, then the voltage
change is -IR.
2. If a resistor is traversed in the direction opposite the current, then the
voltage change is +IR.
3. If a voltage source is traversed from - to +, then the voltage change is +V.
4. If a voltage source is traversed from + to -, then the voltage change is -V.
For a complete loop, the sum of the voltage changes is zero. Now the loop equations and
junction equations can be combined to eliminate unknowns until one quantity can be
determined. Then using that quantity, others can be determined until all the unknowns are
solved for. This is a complete solution of the circuit.
How many loop and junction equations are required for a complete solution of a given
circuit? You need as many loop equations as there are "holes" in the circuit. You need
one fewer junction equations. For example, the circuit in Figure 18.11 of the text has two
holes, so two loop and one junction equations are needed. The circuit in Figure 18.40 has
three holes, so three loop and two junction equations are needed.
Example: P18.16
Figure P18.16 shows a circuit diagram. Determine (a) the current, (b) the potential of
wire A relative to ground, and (c) the voltage drop across the 1500 resistor.
This circuit has one hole, so we will need one loop equation and zero junction equations.
(The green symbol in the bottom left corner is a ground symbol and indicates that that
point in the circuit is at ground potential, or 0V.) Let's label the one current I, and define
the current direction and the direction around the loop to be clockwise.
(a) Because we are always traversing the resistors in the direction of the current, the
voltage change across each will be -IR. The 20V and 25V batteries are traversed from - to
+, so their voltages will enter with a positive sign, but the 30V battery is traversed from +
to -, so its voltage will be given a negative sign. The resulting loop equation is, starting
from the lower left corner:
20.0V -I(2000) -30.0V - I(1000) - I(1500) + 25.0V - I(500) = 0
This can be simplified to:
15.0V -I(5000) = 0 or
I = (15.0V)/(5000) = 0.00300A = 3.00mA
(b) The potential of wire A relative to ground is given by the total change in voltage
when a charge moves from ground to A. Moving up the left side of the circuit from
ground to A, we gain 20.0V across the first battery, then lose -IR = -(3.00mA)(2000) =
-6.0V across the first resistor, then lose -30.0V across the second battery, then lose -IR =
-(3.00mA)(1000) = -3.0V across the second resistor, for a net change of V = 20.0 - 6.0 30.0 - 3.0 = -19.0V.
The voltage drop across the 1500 resistor is IR = (3.00mA)(1500) = 4.5V. The sign of
the voltage drop can only be given if the direction is specified.
Recall:



Torque on a current loop:  = NBIA sin
Galvanometers: based on current loops, have internal resistance and full-scale
current; can be converted to:
o ammeter by the addition of a parallel resistor, or
o voltmeter by the addition of a series resistor.
Charged particle moving perpendicular to a magnetic field follows a circular path
of radius, r = mv/qB.

CH. 19: MAGNETISM
We now begin our study of the second half of "electricity and magnetism". We will learn
about magnetic fields, and how electricity and magnetism are intertwined.
19.1 Magnets
Magnets are common fixtures of everyday life: kitchen magnets, the magnetic stripe on
your student ID card, magnetic hard drives in computers, magnetic sensors of all sorts,
and large magnets used in devices such as MRI. The root of magnetism is derived from
"magnesia", the greek word for the region of Asia Minor where naturally magnetized
magnetite was common.
Magnets have a north (N) and south (S) pole. These opposite poles behave similarly to
plus and minus charges in the case of the electric force. That is, opposite poles attract and
like poles repel. But unlike charges which comes with one sign or the other, magnets
always have both a north pole and a south pole. Even if you cut a magnet in half, each
half will have a north pole and a south pole.
Also, as electric charges have electric fields, magnets have magnetic fields. A picture of
the lines of magnetic field can be created by placing iron filings on paper and placing a
magnet beneath. The iron filings will line up with the magnetic field, making it visible.
course index
Recall from last lecture:



Don't forget Ohm's law: V = I R, we will use this again.
Magnets have north (N) and south (S) poles.
Like poles repel, opposite poles attract.
19.2 Magnetic Field of the Earth
The Earth has a magnetic field of its own. The north pole of a magnet is attracted to the
North magnetic pole of the Earth, and the south pole to the South magnetic pole, hence
the choice of north and south for designating the poles of a magnet.
Since the North magnetic pole attracts the north pole of a magnet, the North magnetic
pole of the Earth is a south pole. And likewise, the South magnetic pole of the Earth is a
north pole. In fact, the magnetic field of the Earth is approximately the field of a large bar
magnet within the Earth, with the south pole of the bar magnet near the North magnetic
pole and the north pole of the magnet near the South magnetic pole.
The North and South magnetic poles aren't located at the North and South poles of the
Earth. (Use globe.) The North magnetic pole is somewhat north of Hudson's Bay in
Canada, about 1300 miles from the North pole.
19.3 Magnetic Fields
A charged particle moving through a magnetic field feels a force given by
F = qvB sin
where q is the charge of the particle, v is the magnitude of the velocity, B is the
magnitude of the magnetic field, and  is the angle between v and B. We turn this
expression around to define the magnetic field:
B = F / (qv sin)
For F in newtons, q in coulombs, v in m/s, the SI unit of magnetic field is the tesla (T).
Flux plays an important role in magnetic fields, and the unit of flux is the weber (Wb).
The relation between magnetic field and flux defines 1T = 1Wb/m². The cgs unit of
magnetic field is also in common use. It is the gauss (G), related to the tesla by 1T =
104G. The Earth's magnetic field is about 0.5G = 0.5×10-4T.
The expression for the force, F=qvBsin, is greatest when sin = 1, or  = 90°. This
corresponds to the charge moving perpendicular to the magnetic field, B. If the charge
moves parallel to the magnetic field, the force is zero.
The direction of the force is perpendicular to both the velocity and the magnetic field.
Use the right hand rule (version #1) to determine the direction of the force:
Hold your right hand open, and orient your fingers in the direction of B (follow the
direction of the arrows if available) and point your thumb in the direction of v (this may
require you to rotate your hand). Now your palm is facing the direction of the force, F,
exerted on a positive charge.
If the charge is negative, then the force is in the opposite direction.
Example: P19.1
An electron gun fires electrons into a magnetic field that is directed straight downward.
find the direction of the force exerted on an electron by the field for each of the following
directions of the electron's velocity: (a) horizontal and due north; (b) horizontal and 30°
west of north; (c) due north, but at 30° below the horizontal; (d) straight upward.
(Remember that an electron has a negative charge.)
Make a sketch of this scenario showing downward magnetic field, and an axis pointing
north. To be sure you can get everything, you may want to draw both a front view and a
side view.
(a) Use the right hand rule, fingers pointing down in the direction of B, thumb pointing
north. Your palm faces west. Since this is a negative charge, the force is directed
oppositely, that is to the east.
(b) Fingers pointing down, thumb pointing 30° west of north. Your palm faces 30° south
of west, so the force is 30° north of east.
(c) Fingers pointing down, thumb pointing north but 30° below the horizontal (bend your
thumb down without moving your fingers). Your palm faces west, so the force is east.
(d) Fingers pointing down, thumb pointing up ... ouch! The pain reminds you that if the
velocity is parallel (or anti-parallel) to the magnetic field then the force is zero. A zero
force has no direction, and that's the answer.
Example: P19.4
A duck flying horizontally due north at 15m/s passes over Atlanta, where the magnetic
field of the Earth is 5.0×10-5T in a direction 60° below a horizontal line running north and
south. The duck has a positive charge of 4.0×10-8C. What is the magnetic force acting on
the duck?
As always, my advice is to begin with a picture. A side view (looking east) should
suffice. Indicate the duck, moving north, and the direction of B (B points northward, at an
angle of 60° from horizontal). Now use F=qvBsin, with  = 60° (your picture should
show that the angle between v and B is 60°):
F = (4.0×10-8C)(15m/s)(5.0×10-5T)sin 60° = 2.6×10-11N. Using the right and rule, fingers
pointing north, 60° below horizontal, thumb pointing north, palms are facing to the west.
Since the charge on the duck is positive, the force points to the west.
19.4 Magnetic Force on a Current-Carrying Conductor
A current is the collective motion of charges. Therefore, if a wire carrying a current is
placed in a magnetic field, the individual charges will feel a force, and they will transmit
this force to the wire. If the wire carries current I, and has length l in a magnetic field, B,
and the current (the wire) makes an angle  with the magnetic field, the resulting force is:
F = BIl sin
The maximum force occurs when the wire is perpendicular to the magnetic field, and is
Fmax = BIl. If the wire is oriented parallel to the magnetic field, then  = 0 and the force is
zero.
19.5 Torque on a Current Loop
Many practical devices involve a loop of wire in a magnetic field. A loop of wire will
experience zero net force, but there will be a net torque. The operation of galvanometers,
generators, and electric motors are based on the torque exerted on a current loop.
Consider a loop of wire in a magnetic field. For simplicity let the loop be rectangular,
with two sides perpendicular to the field, and two sides parallel. The two perpendicular
wires will carry equal currents but in opposite directions, therefore they will feel equal
and oppositely directed forces. The wires parallel to the field feel no force. The net force
is the sum of two equal and opposite forces and this is zero.
Recall from chapter 8 that the torque is defined as  = Fd, where F is the force and d is
the lever arm (the distance perpendicular to the force, from the point of application of the
force to the pivot axis). (The SI unit for torque is newton-meter, Nm.) Let's take the pivot
axis to be through center of the parallel wires, halfway between the applied forces. Then
we see that both applied forces tend to rotate the loop in the same direction, resulting in a
net torque. The general expression we will use for the torque is:
 = NBIA sin
where N is the number of turns of wire in the loop, B is the magnetic field, I is the current
through the wire, A is the are of the loop, and  is the angle between B and a line
perpendicular to the loop. This result applies for a loop of any shape. If required, you
determine the direction of the torque or rotation by remaking a diagram of the loop and
the force that acts on the wire.
Example: P19.22
A 2.00m long wire carrying a current of 2.00A forms a 1 turn loop in the shape of an
equilateral triangle. If the loop is placed in a constant magnetic field of magnitude
0.500T, determine the maximum torque that acts on it.
The maximum torque is max = BIA, where I've set N=1 (1 turn loop) and sin = 1. B and
I are given. The trick here is to determine A from the information given. From what we
are told, we must determine the area of an equilateral triangle whose perimeter is 2.00m.
The length of the sides is equal to one third of the perimeter, or 0.667m. The area is
½base×height, with the base = 0.667m and the height = (0.667m)sin 60° = 0.577m, or A
= ½(0.667m)(0.577m) = 0.192m². Therefore, max = BIA = (0.500T)(2.00A)(0.192m²) =
0.192 Nm.
Recall:




Magnets have north (N) and south (S) poles.
Magnetic field lines run from N to S.
A charge moving in a magnetic field experiences a force F = q v B sin.
Force on a current carrying wire in a magnetic field: F = B I l sin
19.5 Torque on a Current Loop
Many practical devices involve a loop of wire in a magnetic field. A loop of wire will
experience zero net force, but there will be a net torque. The operation of galvanometers,
generators, and electric motors are based on the torque exerted on a current loop.
Consider a loop of wire in a magnetic field. For simplicity let the loop be rectangular,
with two sides perpendicular to the field, and two sides parallel. The two perpendicular
wires will carry equal currents but in opposite directions, therefore they will feel equal
and oppositely directed forces. The wires parallel to the field feel no force. The net force
is the sum of two equal and opposite forces and this is zero.
Recall from chapter 8 that the torque is defined as  = Fd, where F is the force and d is
the lever arm (the distance perpendicular to the force, from the point of application of the
force to the pivot axis). (The SI unit for torque is newton-meter, Nm.) Let's take the pivot
axis to be through center of the parallel wires, halfway between the applied forces. Then
we see that both applied forces tend to rotate the loop in the same direction, resulting in a
net torque. The general expression we will use for the torque is:
 = NBIA sin
where N is the number of turns of wire in the loop, B is the magnetic field, I is the current
through the wire, A is the are of the loop, and  is the angle between B and a line
perpendicular to the loop. This result applies for a loop of any shape. If required, you
determine the direction of the torque or rotation by remaking a diagram of the loop and
the force that acts on the wire.
Example: P19.22
A 2.00m long wire carrying a current of 2.00A forms a 1 turn loop in the shape of an
equilateral triangle. If the loop is placed in a constant magnetic field of magnitude
0.500T, determine the maximum torque that acts on it.
The maximum torque is max = BIA, where I've set N=1 (1 turn loop) and sin = 1. B and
I are given. The trick here is to determine A from the information given. From what we
are told, we must determine the area of an equilateral triangle whose perimeter is 2.00m.
The length of the sides is equal to one third of the perimeter, or 0.667m. The area is
½base×height, with the base = 0.667m and the height = (0.667m)sin 60° = 0.577m, or A
= ½(0.667m)(0.577m) = 0.192m². Therefore, max = BIA = (0.500T)(2.00A)(0.192m²) =
0.192 Nm.
19.6 The Galvanometer and its Applications
A galvanometer makes use of the torque on a current loop placed in the field of a
permanent magnet and can be used to measure current or voltage. A spring is used to
make the deflection proportional to the current in the loop, and a needle is attached to
indicate how far the loop rotates. A typical galvanometer has an internal resistance, r0, of
about 50, and deflects to its maximum for a current of a few milliamps (mA). This is
not directly suitable as a current or voltage measuring device, as an ideal current meter
(ammeter) has zero internal resistance, and an ideal voltmeter has infinite internal
resistance. (Real ammeters have very small, but not zero, internal resistance, and real
voltmeters have very large, but not infinite, internal resistance.) The following example
demonstrates how a galvanometer is made part of an ammeter and voltmeter.
Example: P19.24 (plus extension)
A 50.0, 10.0mA galvanometer is to be converted to an ammeter that reads 3.00A at
full-scale deflection. (a) What value of Rp should be placed in parallel with the coil? (b)
What value of Rs should be placed in series with the coil to convert it to a voltmeter that
reads 10.0V at full-scale deflection?
(a) To convert a galvanometer to an ammeter, we place a resistor in parallel with the coil.
Then part of the current will pass through the coil and the remainder through the parallel
resistor, Rp. The specification of the galvanometer tells us that the internal resistance r0 =
40.0, and that 10.0mA causes the needle to deflect full-scale. Therefore, when 3.00A of
current pass through the coil and resistor in parallel, we want 10.0mA to pass through the
coil, and the remaining 2.99A to pass through Rp. Because the coil and resistor are in
parallel, the same voltage appears across both. When 10.0mA passes through the 50.0
coil, the voltage across it is V = IR = (10.0mA)(50.0) = 500mV = 0.500V. With this
same voltage applied across it, Rp is to carry a current of 2.99A, therefore
Rp = V/I = (0.500V)/(2.99A) = 0.167.
Note that this ammeter does indeed have a small effective resistance of about 0.165.
(b) To convert a galvanometer to a voltmeter, we place a resistor in series with the coil.
The same current passes through the series resistor, Rs, and coil, the added resistance
limits the current through the coil, and increases the effective resistance of the device. If
the needle is to deflect full-scale at a voltage of 10.0V, then the current through resistor
and coil must be 10.0mA. Thus the total resistance of the device is Rtot = Rs + Rcoil = V/I
= (10.0V)/(10.0mA) = 1.00k. Since the coil has a resistance of 50.0, we find
Rs = 1000 - 50.0 = 950.
19.7 Motion of a Charged Particle in a Magnetic Field
When a charged particle moves perpendicularly to a magnetic field, it feels a force F =
qvB directed perpendicular to its motion and the magnetic field. This causes the charge to
change direction, but doesn't affect is speed, v. The charge will follow a circular path.
The force F is directed towards the center of the circle, and equals the centripetal force
needed to keep the charge moving on the circle:
F = qvB = mv²/r
Solving for r gives us an expression for the radius of the circular path in terms of the
magnetic field and the mass, charge, and velocity of the particle:
r = mv/qB
The bending of charged particles in magnetic fields is important in many areas of
Chemistry and Physics. An example is mass spectrometers used to identify particles by
their charge and mass, as described in the example below.
Example: P19.32 mass spectrometer
Consider the mass spectrometer shown schematically in Figure P19.32. The electric field
between the plates of the velocity selector is 950V/m, and the magnetic fields in both the
velocity selector and the devlection chamber have magnitudes of 0.930T. Calculate the
radius of the path in the system for a singly charged ion with mass m=2.18×10-26kg.
The velocity selector is a device which uses both electric and magnetic fields to select
particles moving at particular velocities. Basically, particles can pass through the selector
only if they have a velocity such that the magnetic force balances the electric force:
qvB = qE : or v = E/B.
So, in this device,
v = E/B = (950V/m)/(0.930T) = 1020m/s.
If the particle is singly charged, then q = 1.6×10-19C, and
r = mv/qB = (2.18×10-26kg)(1020m/s)/(1.6×10-19C)(0.930T) = 1.5×10-4m = 0.15mm.
19.8 Magnetic Field of a Long, Straight Wire
We will skip the subsection on Ampère's Law.
A wire carrying a current produces a magnet field. A long straight wire carrying a current
produces a magnetic field that is tangent to a circle centered on the wire and
perpendicular to it. The direction of the magnetic field is given by a second right hand
rule:
If the wire is grasped in the right hand with the thumb in the direction of the current, the
fingers will curl in the direction of B.
The magnetic field a perpendicular distance r from a wire carrying current I is:
B = 0I/2r
The constant 0 is called the permeability of free space and has the value:
0 = 4×10-7Tm/A.
Example: P19.39a
The two wires in Figure P19.39 carry currents of 3.00A and 5.00A in the direction
indicated. Find the direction and magnitude of the magnetic field at a pont midway
between the wires.
Using the right hand rule #2, we find that, halfway between the wires, the magnetic field
produced by the 3.00A wire points straight up and the magnetic field produced by the
5.00A wire points straight down. The midway point is at r=10.0cm from each wire. The
resulting magnetic field is the vector sum of these two contributions:
Btot = B3A - B5A
where define up to be positive.
B3A = 0I/2r = (4×10-7Tm/A)(3.00A)/2(0.100m) = 6.00×10-6T and
B5A = 0I/2r = (4×10-7Tm/A)(5.00A)/2(0.100m) = 10.00×10-6T.
The total field is
Btot = (6.00 - 10.00)×10-6 T = -4.00×10-7T.
The negative sign tells us that the field points downward at that point.
19.9 Magnetic Force Between Two Parallel Conductors
Since a wire carrying a current creates a magnetic field, and a wire carrying a current
feels a force when in a magnetic field, two wires carrying currents will exert a force
between them. Let's consider two parallel conductors carrying currents I1 and I2,
separated by a distance d, and consider the force exerted on wire 1 (draw sketch).
The force on wire 1 will be F1 = B2I1l sin, where B2 is the magnetic field created by wire
2 at the location of wire 1, and  is the angle between wire 1 and the direction of B2.
Wire 2 creates a magnetic field that is tangent circles centered on wire 2. The magnetic
field at wire 1 is tangent to a circle of radius d. Therefore, the lines of magnetic field are
perpendicular to wire 1,  = 90°, and sin = 1. This is true where the conductors are
parallel; if they are at an angle to each other, then the result will be different.
The magnitude of B2 a radius d from wire 2 is B2 = 0I2/2d. Using this in the expression
for the force on wire 1 gives:
F1 = 0I1I2l / 2d.
It is traditional to quite this result as the force per length of parallel wire, that is, to divide
both sides by the length l:
F1/l = 0I1I2 / 2d.
The direction of F1 is determined using the right hand rule -- actually, using both versions
of the right hand rule, first to find the direction of B2, then use that to find the direction of
F1. The force one wire 2 due to wire 1 is equal and opposite to F1, as required by
Newton's third law. These forces tend to either repel or attract the wires from each other,
depending on whether the currents are in the same, or in opposite directions, respectively.
Until recently, this force was used to define the ampere, the unit of current. The coulomb
is then defined as the amount of charge passing a cross section of wire carrying 1A of
current in 1 second.
19.10 Magnetic Field of a Current Loop
Consider again the magnetic field generated by a straight wire, and imagine bending the
wire into a circular loop. The center of the loop is the same distance from any small piece
of wire, and the magnetic field generated by each little piece points in the same direction.
The result is that the field at the center of the loop is enhanced.
19.11 Magnetic Field of a Solenoid
If many loops of wire at stacked together, the field is enhanced even more. A stack of
loops can be made by coiling the wire around a cylinder. Such a configuration is called a
solenoid. A good solenoid is made with many windings of closely spaced wire. The
magnetic field of a good solenoid is particularly simple. To good approximation, the field
is uniform inside the solenoid, directed along the axis of the cylindrical shape, and
outside the solenoid the field is approximately zero (well, much smaller than inside). The
magnitude of the field inside the solenoid is:
B = 0 n I
where n = N/l is the number of turns of wire per unit length of the solenoid, and I is the
current carried in the wire.
Example: P19.47
An electron moves at a speed of 1.0×104m/s in a circular path of radius 2.0cm inside a
solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron's
path. Find (a) the strength of the magnetic field inside the solenoid and (b) the current in
the solenoid if it has 25 turns per centimeter.
(a) To find the strength of the magnetic field, use the result for the radius of the circular
path of a charged particle in a uniform magnetic field: r = m v / q B. Use this to
determine
B = m v / q r = (9.11×10-31kg)(1.0×104m/s) / (1.6×10-19C)(0.020m) = 2.8×10-6T.
(b) Now use the expression for the magnetic field in a solenoid to find the current. We
need the number of turns of wire per meter, n = 25/cm = 2500/m. Since B = 0 n I,
I = B / 0 n = (2.8×10-6T) / (4×10-7Tm/A)(2500/m) = 8.9×10-4A = 0.89mA.
Recall:

Creation of Magnetic Fields:
o Long straight wire: B = 0 I / 2 r
o Solenoid: B = 0 n I



Force Due to Magnetic Field:
o Moving charge: F = q v B sin
o Wire: F = B I l sin
Torque on a Current Loop:  = B I A sin
Force Between Two Wires: F / l = 0 I1 I2 / 2 d
o attractive when the currents are in the same direction
o repulsive when the currents are in opposite directions
CH. 20: INDUCED VOLTAGES AND INDUCTANCE
We have seen that a magnetic field exerts a force on a wire carrying a current, and that a
wire carrying a current generates a magnetic field. Currents are produced by electric
fields, so there seems to be some connection between electricity and magnetism. In this
chapter, we make that connection, seeing how a magnetic field can produce a potential
difference. Magnetic flux will play an important role throughout this chapter.
20.1 Induced emf and Magnetic Flux
Experiments in the 19th century showed that a changing magnetic field can produce an
emf. We quantify the change in terms of magnetic flux. Magnetic flux is defined in a
similar manner to electric flux. For a loop of wire with area A, in a magnetic field, B, the
magnetic flux,  is given by:
 = Bperp A = B A cos
where  is the angle between the perpendicular to the plane of the loop, and the magnetic
field B. The SI units of magnetic flux are Tm².
Another way of looking at flux is to think of it as a count of the magnetic field lines
which pass through the loop. If the loop is oriented perpendicular to the field (=0) then
the flux will be large. If the loop is oriented parallel to the field ( = 90°) no magnetic
field lines pass through the loop, and the flux is zero.
Example: P20.2
A square loop 2.00m on a side is placed in a magnetic field of strength 0.300T. If the
field makes an angle of 50.0° with the normal to the plane of the loop, as in Figure 20.2,
determine the magnetic flux through the loop.
From what we are given, we use
 = B A cos = (0.300T)(2.00m)²cos50.0° = 0.386 Tm²
20.2 Faraday's Law of Induction
Faraday's law of induction relates the change of magnetic flux to induced emf:
emf = -N ( / t)
where N is the number of loops in the coil where the emf is induced, and  is the
change in flux that occurs in time t. This is one of the instances where the term emf
differs somewhat from voltage, and therefore I will use it.
The induced emf is proportional to the change of magnetic flux,  = BAcos. There are
three ways in which the flux can change:
1. a change of the magnitude of the magnetic field, B,
2. a change of the angle between the loop and the magnetic field, , or
3. a change of the area of the loop, A.
The minus sign is there to remind you that the polarity of the induced emf opposes the
change of flux. This is stated precisely in Lenz's law:
The polarity of the induced emf is such that it produces a current whose magnetic field
opposes the change in magnetic flux through the loop. That is, the induced current tends
to maintain the original flux through the circuit.
Example: P20.13
A wire loop of radius 0.30m lies so that an external magnetic field of strength +0.30T is
perpendicular to the loop. The field changes to -0.20T in 1.5s. (The plus and minus signs
here refer to opposite directions through the loop.) Find the magnitude of the average
induced emf in the loop during this time.
The loop is always perpendicular to the field, so the normal to the loop is parallel to the
field, so  = 0, and cos = 1. The flux through the loop is therefore  = BA = Br².
Initially the flux is
i = (0.30T)(0.30m)2 = 0.085Tm²
and after the field changes the flux is
f = (-0.20T)(0.30m)2 = -0.057Tm²
The magnitude of the average induced emf is:
emf = /t = (i - f)/t = (0.085T - (-0.057T))/1.5s = 0.095V = 95mV.
Example: P20.10
The flexible loop in Figure P20.10 has a radius of 12cm and is in a magnetic field of
strength 0.15T. The loop is grasped at points A and B and stretched until it closes. If it
takes 0.20s to close the loop, find the magnitude of the average induced emf in it during
this time.
This is a case where the change in flux is caused by a change in the area of the loop. Both
the magnetic field and the angle  remain constant. When the loop is stretched so that its
area is zero, the flux through the loop is zero. The change in the flux is thus equal to its
original value,
i = B A cos = (0.15T)(0.12m)² = 6.8×10-3Tm²
The average induced emf is thus:
emf = N ( / t) = (6.8×10-3Tm²)/(0.20s) = 3.4×10-2V = 34mV.
20.3 Motional emf
An interesting application of Faraday's law is to produce an emf via motion of the
conductor. As a simple example, let's consider a conducting bar moving perpendicular to
a uniform magnetic field with constant velocity v. For this first look, we have just a bar,
not a complete conducting loop, and we will consider what happens using just the force
on a moving charge, F = qvBsin. This force will act on free charges in the conductor. It
will tend to move negative charge to one end, and leave the other end of the bar with a
net positive charge.
The separated charges will create an electric field which will tend to pull the charges
back together. When equilibrium exists, the magnetic force, F=qvB, will balance the
electric force, F=qE, such that a free charge in the bar will feel no net force. Thus, at
equilibrium, E = vB. The potential difference across the ends of the bar is given by V =
E l, or
V = E l = B l v
This potential difference exists because of the excess of charge at the ends of the
conductor created by motion through the magnetic field. If the direction of motion is
reversed, so is the polarity of the potential difference.
Now let's consider what happens when we add conducting rails for the top and bottom of
the bar to contact, and a resistor between the rails to complete a loop. We can apply
Faraday's law to the complete loop. The change of flux through the loop is proportional
to the change of area from the motion of the bar:
 = B A = B l x.
Using Faraday's law, we find the magnitude of the emf to be (N = 1):
emf = /t = B l x/t = B l v
where I have used the relation v = x/t. This is the same result we obtained considering
the conducting bar by itself.
If the conducting circuit has total resistance, R, then the current is
I = emf / R = B l v / R
Example: P20.18
Over a region where the vertical component of the Earth's magnetic field is 40.0µT
directed downward, a 5.00 m length of wire is held in an east-west direction and moved
horizontally to the north with a speed of 10.0 m/s. Calculate the potential difference
between the ends of the wire, and determine which end is positive.
The vertical component of the magnetic field is perpendicular to the wire and its motion,
so this is what we need. Using the expression obtained for V yields:
V = B l v = (40.0 µT)(5.00 m)(10.0 m/s) = 2.00 mV
To determine which end is positive, consider a positive charge moving north through a
downward magnetic field. The right hand rule gives a force directed to the west. So the
west end of the wire will have a net positive charge, and a more positive potential.
Recall:



Magnetic Flux:  = B A cos
Faraday's law: emf = -N ( / t)
Lenz's law: The induced current tries to keep the magnetic flux through a circuit
constant.
20.3 Motional emf
An interesting application of Faraday's law is to produce an emf via motion of the
conductor. As a simple example, let's consider a conducting bar moving perpendicular to
a uniform magnetic field with constant velocity v. For this first look, we have just a bar,
not a complete conducting loop, and we will consider what happens using just the force
on a moving charge, F = qvBsin. This force will act on free charges in the conductor. It
will tend to move negative charge to one end, and leave the other end of the bar with a
net positive charge.
The separated charges will create an electric field which will tend to pull the charges
back together. When equilibrium exists, the magnetic force, F=qvB, will balance the
electric force, F=qE, such that a free charge in the bar will feel no net force. Thus, at
equilibrium, E = vB. The potential difference across the ends of the bar is given by V =
E l, or
V = E l = B l v
This potential difference exists because of the excess of charge at the ends of the
conductor created by motion through the magnetic field. If the direction of motion is
reversed, so is the polarity of the potential difference.
Now let's consider what happens when we add conducting rails for the top and bottom of
the bar to contact, and a resistor between the rails to complete a loop. We can apply
Faraday's law to the complete loop. The change of flux through the loop is proportional
to the change of area from the motion of the bar:
 = B A = B l x.
Using Faraday's law, we find the magnitude of the emf to be (N = 1):
emf = /t = B l x/t = B l v
where I have used the relation v = x/t. This is the same result we obtained considering
the conducting bar by itself.
If the conducting circuit has total resistance, R, then the current is
I = emf / R = B l v / R
Faraday's law doesn't give us power for free. If the induced emf is used to create a current
in a circuit, then the power supplied to the circuit must come from somewhere. Where? It
comes from the power that must be expended to move the bar through the magnetic field.
The power expended in the circuit is P = I emf = I B l v. By equating this power to the
power expended in moving the bar, we can determine the force required to move the bar.
The work done by an applied force, Fapp, in moving the bar a distance x is W = Fapp x.
If this takes time t, the power delivered by the applied force is P = W / t = Fapp x / t
= Fapp v. Equating the power delivered with the power expended gives:
Fapp = I B l
This force is the same magnitude as the force exerted on the bar by the magnetic field due
to the current through the bar:
F = B I l sin = B I l.
The direction of the applied force will be opposite the direction of the magnetic force on
the bar, as we will discuss more in the next section. This result demonstrates that the
results in this chapter and the last are consistent, that is, everything ties together, and that
often there is more than one way in which to view a particular problem.
Example: P20.18
Over a region where the vertical component of the Earth's magnetic field is 40.0µT
directed downward, a 5.00 m length of wire is held in an east-west direction and moved
horizontally to the north with a speed of 10.0 m/s. Calculate the potential difference
between the ends of the wire, and determine which end is positive.
The vertical component of the magnetic field is perpendicular to the wire and its motion,
so this is what we need. Using the expression obtained for V yields:
V = B l v = (40.0 µT)(5.00 m)(10.0 m/s) = 2.00 mV
To determine which end is positive, consider a positive charge moving north through a
downward magnetic field. The right hand rule gives a force directed to the west. So the
west end of the wire will have a net positive charge, and a more positive potential.
Lenz's Law Revisited
Application of Lenz's law will tell us the direction of induced currents, the direction of
applied or produced forces, and the polarity of induced emf's. The idea is to determine the
direction of change of the magnetic flux, then find the direction of induced current that
will produce magnetic flux in the opposite direction. Up to now we haven't really spoken
of a direction for magnetic flux, but since it is derived from the magnetic field lines, and
they have a direction, we can use the direction of the lines. If the magnitude of the flux is
increasing, then the change is in the direction of the magnetic field (the part going
through the loop). If the magnitude of the flux is decreasing, then the change is opposite
the direction of the magnetic field.
For instance, consider again the bar moving along rails through a magnetic field. As the
bar moves to the right, more magnetic field lines pass through the loop, so the magnitude
of the flux is increasing. We say it is increasing into the page (the magnetic field is into
the page).
Lenz's law says that the induced current will produce magnetic flux opposing this change.
To oppose an increase into the page, it generates magnetic field which points out of the
page, at least in the interior of the loop. Such a magnetic field is produced by a
counterclockwise current (use the right hand rule to verify).
If instead the bar is moving to the left, then the magnitude of the flux is decreasing as the
area of the loop decreases. The change of flux is directed opposite the magnetic field, or
out of the page. The induced current will produce a magnetic field opposing the change,
in this case a magnetic field directed into the page. And using the right hand rule, this
requires a clockwise current. We reverse the motion of the bar and the direction of the
induced current is also reversed.
Example: P20.23
In Figure P20.23, what is the direction of the current induced in the resistor at the instant
the switch is closed.
To solve problems of this type, involving Lenz's law, doesn't require calculations, but it
does require an understanding of the right hand rule, and careful consideration of the
direction of change of magnetic flux. In this problem, before the switch is closed, there is
no current in the solenoid, and no magnetic field inside. After the switch is closed,
current will flow and a magnetic field is created in the solenoid (you can assume that the
field outside of the solenoid is zero). The loop containing the resistor encloses the
solenoid, so the magnetic field will pass through the loop, that is, there is magnetic flux
through the loop after the switch is closed. The change from zero flux to non-zero flux
will produce a current in the loop and passing through the resistor. Now we must
determine the direction of that current.
First, let's find the direction of the current through the solenoid. The current will flow
from the positive battery terminal to the negative, so it will flow from the battery, through
the switch, then through the solenoid, and back to the battery. The resulting magnetic
field is directed to the left. The induced current in the loop with resistor will produce a
magnetic field directed to the right. That means the current through the resistor will flow
left to right.
Notice that if we now ask for the direction of the current when the switch is opened
again, we will get the opposite answer. That is, the magnetic field of the solenoid is
initially pointing to the left, and goes to zero when the switch is opened. The magnitude
of the magnetic flux is decreasing, so the change in the flux is opposite the direction of
the field (the field that exists during the change). The change in the flux is directed to the
right. The induced current in the loop with resistor will produce a magnetic field directed
to the left, and therefore the current through the resistor is from right to left.
While the switch is not moved, remaining either opened or closed, the magnetic flux is
not changing, and there is no current in the loop with resistor.
Recall from last lecture:



Magnetic flux:  = B A cos
Faraday's law: emf = -N  / t
Lenz's law: The induced current opposes the change of magnetic flux.
20.5 Generators and Motors
Generators and motors are two of the most important applications of induced emf
(magnetic inductance). A generator is something that converts mechanical energy to
electrical energy. A motor does the reverse; it converts electrical energy to mechanical
energy. Here I'll discuss two types of generators, and say a few words about motors.
The ac generator
The basic generator design consists of a loop of wire capable of rotating inside a uniform
magnetic field (sketch similar to Figure 20.17). Let's determine the emf generated as the
loop rotates. This discussion differs from the one presented in the text, to give you an
alternative way of thinking about the situation.
The loop has dimensions a by l, and area A = al; the magnetic field is of magnitude B,
and directed horizontally from left to right. Now we assume the loop is rotating with
constant angular speed , or frequency f = 2. When the loop is at an angle  = t
from vertical, the flux through the loop is i = B A cos(t), since  = t is the angle
between B and the normal to the plane of the loop. This means that a time t later, the
loop is at an angle ' (t + t). The flux through the loop is now f = B A cos(t + t).
The change in flux is
 = f - i = B A {cos(t + t) - cost}.
The induced emf is given by Faraday's law:
emf = -N / t = N B A  sint,
where I obtain the final form using the rules of calculus. The factor N is left in to leave
the result in a form correct when the loop consists of more than one turn of wire.
In a practical generator, the emf is conveyed out of the coil to an external circuit using a
pair of rings and brushes (diagram). The resulting emf varies sinusoidally with time
(sketch). This is type of voltage is called alternating current (ac). The voltage varies
between a maximum value of emfmax = NBA, and a minimum of emfmin = -NBA.
The dc generator
By a clever change to the rings and brushes of the ac generator, we can create a dc
generator, that is, a generator where the polarity of the emf is always positive. The basic
idea is to use a single split ring instead of two complete rings. The split ring is arranged
so that, just as the emf is about to change sign from positive to negative, the brushes cross
the gap, and the polarity of the contacts is switched. The polarity of the contacts changes
in phase with the polarity of the emf -- the two changes essentially cancel each other out,
and the emf remains always positive. The emf still varies sinusoidally during each half
cycle, but every half cycle is a positive emf (drawing, like Figure 20.18(b)).
Motors
A motor is basically a generator running in reverse. A current is passed through the coil,
producing a torque and causing the coil to rotate in the magnetic field. Once turning, the
coil of the motor generates a back emf, just as does the coil of a generator. The back emf
cancels some of the applied emf, and limits the current through the coil.
Example: P20.31
A coil of area 0.10 m² is rotating at 60 rev/s with its axis of rotation perpendicular to a
0.20T magnetic field. (a) If there are 1000 turns on the coil, what is the maximum voltage
induced in the coil? (b) When the maximum induced voltage occurs, what is the
orientation of the coil with respect to the magnetic field?
(a) First we need the angular speed  = 2f = w(60 Hz) = 377 rad/s. emfmax = N B A 
= (1000)(0.20T)(0.10m²)(377 rad/s) = 7500V.
(b) When the maximum induced voltage occurs, sint = 1.0, so t = 90°. This is the
angle between the normal to the loop and the magnetic field, so the plane of the coil is
parallel to the magnetic field.
Recall from last lecture:

Generators: emf = N B A  sin(t)
20.7 Self-Inductance
Consider a circuit consisting of a solenoid, a resistor, a battery, and a switch (sketch
circuit). If initially the switch is open, then no current flows in the circuit and the
magnetic field (and flux) in the solenoid is zero. When the switch is closed, current
begins to flow through the circuit. The current in the solenoid produces magnetic flux in
the solenoid. According to Lenz's law, this increase in flux will be opposed by an induced
emf (and corresponding current) in the windings of the solenoid. This opposing emf is
such as to limit the current to less than its nominal value of i0 = V/R. A short instant later,
the induced emf will abate, and the current through the solenoid can increase. The
increase in current produces another increase in flux, and a corresponding opposing emf.
This process continues until the current reaches the nominal value. This phenomenon is
known as self-induction because the changing flux through a circuit comes from the
circuit itself.
Now let us look at the situation more quantitatively. The induced emf is given by
Faraday's law:
emf = -N  / t.
Let's assume that we have an ideal solenoid, in which case the magnetic field inside is B
= 0nI, and the flux through the solenoid is  = BA = 0nIA, where A is the crosssectional area of the solenoid. Any change in flux must be produced by a change in the
current, I, since 0, n, and A remain constant. Therefore,  = 0n A I. Going back to
the expression for the induced emf, we have:
emf = -N 0 n A I / t = -L I / t
where L is a proportionality constant called the inductance of the device. This result says
that the induced emf is proportional to the time rate of change of the current.
The inductance is the proportionality constant between the time rate of change of the
current and the induced emf. Note the minus sign in the expression for self-induced emf.
This means that if the current is increasing, the emf is negative, so as to oppose the
increase, and if the current is decreasing, the emf is positive, opposing the decrease.
The unit of inductance is called the henry, abbreviated H. From the expression for selfinduced emf we see that 1H = 1Vs/A. In the special case of an ideal solenoid,
L = 0 N n A.
Recall that n = N/L, so we can write the inductance of an ideal solenoid as
L = 0 N² A / L.
This is the analog of the expression for the capacitance of a parallel plate capacitor, C =
0A/d.
Example: P20.36
A solenoid of radius 2.5cm has 400 turns and a length of 20 cm. Find (a) its inductance
and (b) the rate at which current must change through it to produce an emf of 75mV.
(a) Using the expression for the inductance of a solenoid:
L = 0 N² A / L = 4×10-7 (400)²  (0.025m)² / 0.20m = 2.0×10-3H = 2.0mH.
To induce an emf of 75mV, the current must change at the rate (ignore the minus sign for
this calculation, it only effects whether the current increases or decreases, but the value of
the rate remains the same)
I/t = emf/L = (75mV)/(2.0mH) = 38 A/s.
20.8 RL Circuits
There are many parallels between inductance and capacitance. We've already seen the
parallel between the expressions for the capacitance of a parallel plate capacitor and the
inductance of a solenoid, and though they may seem unrelated, the relations between
voltage and charge for a capacitor, Q = CV, and between induced emf and rate of change
of current for an inductor, emf = -L I / t.
Now let's consider the parallel of the capacitor, the inductor. An inductor is a circuit
element having inductance, L, and is represented in circuit schematics with a symbol that
looks like a coil of wire (sketch).
A circuit containing a resistor and inductor is called an RL circuit. Let's look at a circuit
containing a resistor, R, an inductor, L, a battery, V, and switch, and ask what happens
when the switch is closed. While the switch is open, no current can flow through the
circuit, and there is no magnetic field in the inductor. When the switch is closed, current
begins to flow, magnetic flux is created in the inductor, and an emf is setup opposing the
flux. Due to the resistance, the maximum current that can flow is I0 = V/R. The actual
current flow is initially less, due to the presence of the inductor:
I = (V/R)(1 - e-Rt/L) = I0 (1 - e-t/)
where  = L/R is the time constant for the circuit, similar to the time constant for RC
circuits.
20.9 Energy Stored in a Magnetic Field
There is yet another similarity between capacitors and inductors, they both store energy.
Recall that the energy stored in a capacitor with voltage V between its plates, holding
charge Q is:
PEC = ½QV = ½CV² = Q²/2C
We find it useful to think of this energy as being stored in the electric field created
between the plates of the capacitor. Inductors don't have electric fields, but there is a
magnetic field, and it takes energy to create. An inductor L, carrying current I, holds
energy equal to:
PEL = ½LI²
Example: P20.45
A 24V battery is connected in series with a resistor and an inductor, where R = 8.0 and
L = 4.0H. Find the energy stored in the inductor (a) when the current reaches its
maximum value and (b) one time constant after the switch is closed.
(a) The maximum value of the current is I0 = V/R = 24V / 8.0 = 3.0A. The energy
stored in the inductor is then:
PEL = ½LI² = (4.0H)(3.0A)²/2 = 18J
(b) One time constant after the switch is closed (that is, after current begins flowing in the
circuit), I = I0(1 - e-1) = 0.37 I0 = 1.11A. Then energy stored at this current is:
PEL = ½LI² = (4.0H)(1.11A)²/2 = 2.5J.
Summary of the properties of circuit elements.
Resistor
Capacitor
Inductor
units
farad, F = C / V
henry, H = V s / A
ohm,  = V / A
symbol
R
C
L
relation
V=IR
Q=CV
emf = -L (I / t)
power dissipated
P = I V = I² R = V² / R
0
0
energy stored
0
PEC = C V² / 2
PEL = L I² / 2
Recall:
Summary of the properties of circuit elements.
Resistor
Capacitor
Inductor
units
ohm,  = V / A
farad, F = C / V
henry, H = V s / A
symbol
R
C
L
relation
V=IR
Q=CV
emf = -L (I / t)
power dissipated
P = I V = I² R = V² / R
0
0
energy stored
0
PEC = C V² / 2
PEL = L I² / 2
time constant
0
 = RC
 = L/R
CH. 21: ALTERNATING CURRENT CIRCUITS AND
ELECTROMAGNETIC WAVES
This is the final chapter in our discussion of electricity and magnetism. We will begin the
chapter with an examination of ac circuits; simple circuits consisting of resistors,
capacitors, and inductors connected to a source of ac voltage, like the generator of the last
chapter. AC circuits are important, because our electrical power grid is based on
delivering power via ac currents.
The latter half of the chapter introduces the concepts and equations related to
electromagnetic waves. Electromagnetic waves span the spectrum from micro waves to
radio waves to infrared radiation to visible light to ultraviolet light to X-rays and gamma
rays.
21.1 Resistors in an ac Circuit
An ac circuit is a circuit powered by an ac generator. We discussed ac generators in
chapter 20, and for the purposes here, they can be considered as a voltage source that
varies sinusoidally with time (called an ac voltage source):
v = Vm sin(t) = Vm sin(2 f t)
where Vm is the maximum voltage that appears, and f is the frequency in hertz. In this
chapter, I'll denote voltages and currents that vary with time by lower case letters, v and i.
Upper case letters, like Vm, represent constant voltage or current values.
Consider a circuit consisting of a resistor and an ac voltage source (sketch, showing
symbol for ac voltage source). What is the current through the resistor? At any instant,
the voltage across the resistor is v, and the current is i = v/R. The result is as shown in
Figure 21.2:
i = v/R = (Vm / R) sin( 2 f t) = Im sin( 2 f t)
When plotted as a function of time, we see that the current is always proportional to the
voltage, that is, it varies sinusoidally, and remains in phase with the voltage. When the
voltage is positive the current is positive; when the voltage reaches its peak, the current
reaches its peak; when the voltage is zero, the current is zero.
The average value of the current over one cycle is zero. (The same is true of the voltage.)
This does NOT mean that the power dissipated in the resistor is zero. The current is zero
because it is positive for half a cycle, then negative for the other half. But during each
half, the power dissipated in the resistor is positive; a current moving through a resistor
dissipates power, no matter which way it is moving.
It is useful to have an expression for the (average) power dissipated. Start with the
expression for power:
P = i² R = Im² R sin²( 2 f t)
Next, we must average this expression over one cycle. Since Im and R are constants, the
only thing to average is the sin² function, and the average of sin² over one cycle is 1/2.
Thus:
P = ½ Im² R
The average current squared we call Ir² = <i²>av = ½ Im².
Ir = Im / sqrt(2) = 0.707 Im
The square root of this current, Ir, we call the root mean square (rms) current. The rms
current is equal to the amount of direct current that will dissipate the same energy as the
alternating current:
P = Ir² R
Similarly, we define the rms voltage as:
Vr = Vm / Sqrt(2) = 0.707 Vm
The 120V from our electrical outlets is the rms voltage. The peak voltage is about 170V.
In terms of rms voltage and current, we can write the ac equivalent of Ohm's law:
Vr = Ir R
Example: P21.2
An ac voltage source has an output of v = 150 sin 377t. Find (a) the rms voltage output,
(b) the frequency of the source, and (c) the voltage at t = (1/120)s. (d) Find the maximum
current in the circuit when the generator is connected to a 50.0 resistor.
(a) Use the relation between rms and peak voltage. The peak voltage is the value when
the sine function is 1.
Vr = Vm / sqrt(2) = 0.707(150V) = 106V.
(b) The operand of the sine function equals 2 f t, therefore,
f = 377/2 = 60.0 Hz.
(c) At t=(1/120)s,
v = 150 sin (377/120) = 0.V
Remember that the operand of the sine function is in radians, so set your calculator to
radians before calculating.
(d) When connected to a 50.0 resistor, the maximum current is related to the maximum
voltage via Ohm's law:
Im = Vm / R = 150V / 50.0 = 3.00A.
21.2 Capacitors in an ac Circuit
Next, we want to understand how a capacitor affects an ac circuit, so let's begin with the
simplest circuit, just a capacitor and ac voltage source (sketch). Consider what happens as
the voltage goes through a cycle (Figure 21.5). First, the capacitor is uncharged, so as the
voltage increases, current flows easily to begin charging the capacitor. Then, as the
voltage increases, more charge is added, but at a slower rate, so the current decreases.
When the voltage reaches its peak value, current momentarily stops flowing. As the
voltage decreases, the capacitor begins to discharge, and the current becomes negative.
And you get the idea.
The capacitor limits the current in the current. In addition, the current and voltage are out
of phase; we say that the voltage lags the current, since the current peaks a quarter of a
cycle (90°) before the voltage peaks.
We summarize the effect of the capacitor by the capacitive reactance:
XC = 1 / (2 f C)
The reactance has units of ohms, just like resistance. In terms of the reactance, the rms
voltage across the capacitor is related to the rms current through the capacitor by:
VC = Ir XC
Recall:



An ac voltage source has a sinusoidally varying output: v = Vmsin 2 f t
RMS voltage and current: Vr = vm / sqrt(2), Ir = Im / sqrt(2)
See the summary after section 21.3.
21.3 Inductors in an ac Circuit
And finally, consider what happens when an inductor is put in a circuit with an ac voltage
source. Similar arguments apply as for the capacitor, but this time we find that the current
leads the voltage by a quarter cycle (90°), and that the rms current is related to the rms
voltage across the solenoid by:
VL = Ir XL
The quantity XL is called the inductive reactance, given by:
XL = 2 f L
Example: P21.15
A 2.40F capacitor is connected across an alternating voltage with an rms value of
9.00V. The rms current in the capacitor is 25.0mA. (a) What is the source frequency? (b)
If the capacitor is replaced by an ideal coil with an inductance of 0.160H, what is the rms
current in the coil?
(a) From the rms voltage and current, we can determine the capacitive reactance, and this
is related to the capacitance and the frequency of the source.
XC = Vr / Ir = 9.00V / 25.0mA = 360.
f = 1 / (2 C XC) = 1 / (2(2.40F)(360)) = 184Hz.
At this frequency, the inductive reactance is
XL = 2 f L = 2 (184Hz)(0.160H) = 185.
The rms current can be found from the rms voltage and reactance:
Ir = Vr / XL = 9.00V / 185 = 48.6mA.
Summary of the properties of ac circuit elements.
Resistor
Capacitor
Inductor
units
ohm,  = V / A
farad, F = C / V
henry, H = V s / A
symbol
R
C
L
reactance
XR = R
XC = 1 / (2 f C)
XL = 2 f L
phase
0°
-90°
90°
21.4 The RLC Series Circuit
Now we want to examine what happens when a resistor, capacitor, and inductor are
combined in an ac circuit. We begin by considering the circuit consisting of an ac voltage
source, a resistor, a capacitor, and an inductor in series (sketch circuit like Figure 21.8).
What should we expect to happen in this circuit, that is, what will be the relation between
voltage and current? This is a series circuit, so the current will be the same through all the
elements. Although the current is the same, the voltages across each element will in
general be different in magnitude and in phase. The result is that the voltage from the ac
source need not be in phase with the current, and the relative phase can be essentially
anything.
Now let's work this out quantitatively. If we let the current be
i = Im sin 2 f t
then the voltage drop across the resistor is
VR = i R,
the voltage drop across the capacitor is
VC = i XC,
and the voltage drop across the inductor is
VL = i XL.
Note that the frequency of the current is the same as the frequency of the voltage source.
From the previous sections we know that the voltage drop across the resistor is in phase
with the current, the voltage drop across the capacitor lags the current by 90°, and the
voltage drop across the inductor leads the current by 90°. We can represent all of this
with a phasor diagram. A phasor diagram is like a vector diagram, but representing
voltages (or reactances) and the relative phases between them (draw phasor diagram for
voltages). From this, we get the following triangle for the voltage sum (diagram like
Figure 21.11). The hypotenuse of the triangle is the total voltage of the voltage source,
and the angle it makes with the x-axis is the phase of the voltage with respect to the
current:
V = Sqrt{VR² + (VL - VC)²}
tan  = (VL - VC) / VR
Because the voltages are related to the reactances by a common factor, the current, we
can also make a phasor diagram for the reactances. The hypotenuse of the reactance
diagram is called the impedance,
Z = Sqrt(R² + (XL - XC)²}.
tan  = (XL - XC) / R
In terms of the impedance, we can write Ohm's law for ac circuits:
V = I Z
where V and I can be either rms values, or peak values, but don't mix them and choose
one of each!
Example: P21.20
A 50.0 resistor, a 0.100H inductor, and a 10.0F capacitor are connected in series to a
60.0Hz source. The rms current in the circuit is 2.75A. Find the rms voltages across (a)
the resistor, (b) the inductor, (c) the capacitor, and (d) the RLC combination. (e) Sketch
the phasor diagram for this circuit.
Since the rms voltage is related to the peak voltage by a constant, we can use the Ohm's
law relations for rms voltages and currents.
(a) VR(rms) = Ir R = (2.75A)(50.0) = 138Vrms.
(b) XL = 2 f L = 2 (60.0Hz)(0.100H) = 37.7
VL(rms) = Ir XL = (2.75A)(37.7) = 104Vrms.
(c) XC = 1/2 f C = 1 / (2(60.0Hz)(10.0F) = 265
VC(rms) = Ir XC = (2.75A)(265) = 729Vrms
(d) The rms voltage drop across the RLC combination can be determined using the
expression for the hypotenuse of the voltage triangle.
V = Sqrt{VR² + (VL - VC)²} = Sqrt{(138V)² + (104V - 729V)²} = 640V.
21.5 Power in an ac Circuit
As mentioned previously (see the summary table at the beginning of section 21.1),
capacitors and inductors store energy, but do not dissipate energy. Only resistors dissipate
energy. Therefore, the power dissipated in an ac circuit depends only on the resistance in
the circuit
Pav = I² R.
Since R = VR / I, the average power can also be written:
Pav = I VR.
Finally, we can express VR = V cos, as can be seen from the voltage triangle of the
previous section. This gives us a third form for the average power:
Pav = I V cos. The cos  factor is known as the power factor.
Example: P21.26
In a certain RLC circuit, the rms current is 6.0A, the rms voltage is 240V, and the current
leads the voltage by 53°. (a) What is the total resistance of the circuit? (b) Calculate the
total reactance, XL - XC. (c) Find the average power dissipated in the circuit.
Begin by drawing a phasor diagram of the rms voltage. Because the current leads the
voltage, the voltage lags the current, so the phase angle is  = -53°. (a) VR is the
component of the voltage along the x-axis, and is related to the resistance by R = VR / Ir.
VR = V cos = (240V) cos-53° = 144V
R = VR / Ir = (144V) / (6.0A) = 24
(b)XL - XC = R tan = 24 tan-53° = -32
(c) Pav = I V cos = (6.0A)(240V)cos-53° = 870 W.
21.6 Resonance in a Series RLC Circuit
Recall:


The summary table of ac properties of circuit elements.
Phasor diagrams: used for finding the voltage-current relation in an ac circuit.
o Voltage: V = Sqrt{VR² + (VL - VC)²}
o Impedance: Z = Sqrt{R² + (XL - XC)²}
o Phase: tan = (VL - VC) / VR = (XL - XC) / R
21.5 Power in an ac Circuit
As mentioned previously (see the summary table at the beginning of the lecture of Oct.
20, 2000), capacitors and inductors store energy, but do not dissipate energy. Only
resistors dissipate energy. Therefore, the power dissipated in an ac circuit depends only
on the resistance in the circuit
Pav = I² R.
Since R = VR / I, the average power can also be written:
Pav = I VR.
Finally, we can express VR = V cos, as can be seen from the voltage triangle of the
previous section. This gives us a third form for the average power:
Pav = I V cos. The cos  factor is known as the power factor.
Example: P21.26
In a certain RLC circuit, the rms current is 6.0A, the rms voltage is 240V, and the current
leads the voltage by 53°. (a) What is the total resistance of the circuit? (b) Calculate the
total reactance, XL - XC. (c) Find the average power dissipated in the circuit.
Begin by drawing a phasor diagram of the rms voltage. Because the current leads the
voltage, the voltage lags the current, so the phase angle is  = -53°. (a) VR is the
component of the voltage along the x-axis, and is related to the resistance by R = VR / Ir.
VR = V cos = (240V) cos-53° = 144V
R = VR / Ir = (144V) / (6.0A) = 24
(b)XL - XC = R tan = 24 tan-53° = -32
(c) Pav = I V cos = (6.0A)(240V)cos-53° = 870 W.
21.6 Resonance in a Series RLC Circuit
Resonance is a phenomenon where system responds strongly to an exciting force with a
particular frequency. For instance, the string of a guitar or violin vibrates at a particular
frequency when plucked, the air in a wind instrument when excited, the tines of a tuning
fork, or the crystal used to keep time in an electronic (quartz) watch. Resonance also
occurs in RLC circuits.
If the ac voltage source has a peak value Vm, then the peak current can be written:
Im = Vm / Z = Vm / {R² + (XL - XC)²}
Remember that XL and XC depend on the frequency of the source. We will get a
maximum current in the circuit when XL = XC, so that Z = R. This occurs at the
resonance frequency, f0, where
2 f0 L = 1 / 2 f0 C
From this we find
f0 = 1 / 2 Sqrt{LC}.
Show a plot of the current versus frequency, as in Figure 21.23.
Note that if the resistance of the circuit, R, is zero, then the current becomes infinite at
resonance. The small resistance in real devices keeps this from happening.
Applications of resonance include receivers for television and radio, and the detector in
an MRI. In fact, the R in MRI stands for resonance -- magnetic resonance imaging.
Example: P21.29
An RLC circuit is used to tune a radio to an FM station broadcasting at 88.9MHz. The
resistance in the circuit is 12 and the capacitance is 1.40pF. What inductance should be
present in the circuit?
The resistance given is not important and will not be used. We want the resonance
frequency of the circuit to equal the frequency of the station, and this depends only on the
inductance and capacitance:
f0 = 1 / 2 Sqrt{LC}
Therefore, solving for L:
L = 1 / 4² f0² C = 1 / 4² (88.9×106Hz)² (1.40×10-12F) = 2.29×10-6H = 2.29H.
21.11 Properties of Electromagnetic Waves
I'll begin this section with some introduction from sections 21.8, 21.9, and 21.10. Since
electric and magnetic fields are very similar in behavior, and, as we've seen, changing
magnetic fields produce electric fields (seen by us as emf's), one might wonder if
changing electric fields produce magnetic fields. In 1865, Maxwell proposed just this.
When he included this effect among the equations of electricity and magnetism, he
noticed that the resulting equations also allowed waves to exist, and that the speed of
these waves predicted by the equations was very close to the speed of light! (speed of
light = c = 3×108 m/s) He concluded, and it was later verified, that light is an
electromagnetic wave, that is a wave made by fluctuating electric and magnetic fields.
Experiments were carried out by Hertz that confirmed the existence of electromagnetic
waves, and demonstrated that his waves (f = 100MHz) had properties very similar to light
waves (f = 500THz). These properties include a propagation speed equal to the speed of
light, reflection, refraction, interference, and polarization. In Maxwell's theory, the speed
of electromagnetic waves is
c = 1 / Sqrt{0 0}
If we substitute the values 0 = 4×10-7 N s² / C² and 0 = 8.85×10-12 C² / N m² we find c
= 2.998×108 m/s. Compare this with the exact value for c as defined by international
agreement: c = 2.99792458×108 m/s (exactly).
Maxwell's theory also gives us the structure of electromagnetic waves. Electromagnetic
waves consist of fluctuating electric and magnetic fields, with both fields perpendicular
to the direction of motion and perpendicular to each other. Because the electric and
magnetic fields are perpendicular (transverse) to the motion, electromagnetic waves are
transverse waves. The ratio of the magnitude of the electric and magnetic fields equals
the speed of light:
E / B = c.
Electromagnetic waves carry energy and momentum! The average power per unit area
transferred by an electromagnetic wave is
Pav / A = Em Bm / 20 = Em² / 20c = c Bm² / 2 0
The various forms are equivalent due to the relation E / B = c. The amount of momentum
transferred to an absorbing surface is related to the energy U incident on the surface:
p = U / c.
If the surface is perfectly reflecting, then the momentum transfer is double the above
value:
p = 2U / c.
21.12 The Spectrum of Electromagnetic Waves
Radio waves, microwaves, infrared waves, visible light, ultraviolet light, x-rays, and
gamma rays are all electromagnetic waves. They differ only in their frequency, f, or
wavelength, . Because all electromagnetic waves travel (through vacuum) at the speed
of light, c, the frequency and wavelength are related through:
c=f
Figure 21.23 shows the frequency and wavelength for different classes of electromagnetic
waves.
It is traditional to use frequency for waves with wavelength greater than about 1 micron.
These are the microwaves and radiowaves.
For waves with wavelengths less than 1 micron, it is more common to specify the
wavelength. For instance, the wavelength of visible light lies in the range from 700nm
(red) to 400nm (blue).
© Robert Harr 2000