Download R - Uplift North Hills Prep

Charge – the conservation of charge
PRACTICE: A cat’s fur acts like your hair, when you
rub a balloon on it. A balloon has picked up
-150 C of charge from Albert. The resulting
opposite charges of balloon and Albert cause the
balloon to stick.
(a) How many electrons have been transferred?
(b) What is the charge on Albert?
SOLUTION:
1 e- = -1.6010-19 C so…
(a) n = (-15010-6 C)/ (-1.6010-19 C) = 9.41014 e-.
(b) Conservation of charge tells us that qAlbert = +150 C.
PRACTICE: Many houses have 20-amp(ere) service. How many electrons per
second is this?
SOLUTION:
20 A is 20 C (per s) so we only need to know how many electrons are in 20 C.
(20 C)(1 e- / 1.610-19 C) = 1.31020 e-.
EXAMPLE - QUESTION Charging by induction
Bring a charged object near a conducting surface, electrons will move in conductor even
though no physical contact: Due to attraction or repulsion of electrons in conductor to the
charged object – since free to move, they will!
Once separated from each other with rod still close they’ll remain charged. Charge is
conserved, so charges on spheres A and B are equal and opposite.
Note, the charged rod never touched them, and retains its original charge.
Charge – detection using an electroscope
PRACTICE: Consider the three electroscopes shown.
Which one has the greatest charge in the leaves?
Which one has the least?
Can you tell whether the charge is (+) or (-)? Why?
SOLUTION:
The last one has the
most charge, the middle
one the least.
You cannot tell the
sign of the charge
since (-)(-) will repel,
but so will (+)(+).
during
PRACTICE: Explain: A charged wand is brought near
an uncharged electroscope without touching it.
While the wand is near, the leaves spread apart.
SOLUTION:
The ball and the leaves are conductors
and they are connected to each other.
The wand’s charge repels like charges
in the ball.
The like charges in the ball travel as
far as they can to the leaves.
The leaves now temporarily hold like
charges and thus they repel each other.
before and
after
Charge – detection using an electroscope
Practice: Coulomb’s law
Find the Coulomb force between two electrons located 1.0 cm apart.
SOLUTION:
Note r = 1.0 cm = 0.010 m.
Note q1 = e = 1.6010-19 C.
Note q2 = e = 1.6010-19 C.
From F = kq1q2 / r 2
F
= 8.99109(1.6010-19)2 / 0.0102
= 2.3010-24 N.
Since like charges repel the electrons repel.
PRACTICE: Find the electric field strength 1.0 cm from an electron.
SOLUTION: We have already found the Coulomb force between two electrons
located 1.0 cm apart. We just divide our previous answer by one of the charges:
Then E = F / q = 2.310-24 / 1.610-19 = 1.410-5 N C-1.
SHE accumulates a charge q1 of 2.0 x 10-5 C
(sliding out of the seat of a car).
HE has accumulated a charge q2 of – 8.0 x 10-5 C
while waiting in the wind.
What is the force between them
a) when she opens the door 6.0 m from him and
b) when their separation is reduced by a factor of 0.5?
a) They exert equal forces on each other only in opposite direction
F k
q1q2
 0.40 N
2
r
(“-“ = attractive force)
b) r’ = 0.5 r
q1q2
F '  k 2  1.6 N  4 F
r'
Strong force at very small separation
How many electrons is 2.0 x 10-5 C ?
2.0 10 5 C
14

10
electrons
19
1.6 10 C
spark
Three point charges : q1= +8.00 mC; q2= -5.00 mC and q3= +5.00 mC.
(a) Determine the net force (magnitude and direction) exerted on q1 by the
other two charges.
(b) If q1 had a mass of 1.50 g and it were free to move, what would be its
acceleration?
1.30 m
230
q1
q2
230
1.30 m
Force diagram
F3
q3
q1
F2
F2  k
q1q2
 0.213 N
2
r
q1q2
F3  k 2  0.213 N
r
𝐹𝑥 = 0 , 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑓 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦; x-components will cancel each other
𝐹𝑦 = 𝐹2 sin 230 + 𝐹3 sin 230 = 0.213 sin 230 + 0.213 sin 230 = 0.166 𝑁
F = Fy = 0.166 N
𝑎=
𝐹
0.166
𝑚
=
=
111
𝑚 1.50 × 10−3
𝑠2
𝑖𝑛 𝑦 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
electric force is very-very strong force, and resulting acceleration can be huge
A positive and negative charge with equal magnitude are connected by a rigid rod,
and placed near a large negative charge. In which direction is the net force on the
two connected charges?
1) Left
2) Zero
3) Right
Positive charge is attracted (force to left)
Negative charge is repelled (force to right)
Positive charge is closer so force to left is larger.
-
+
-
Coulomb’s law – permittivity – practice
If the two electrons are embedded in a chunk of quartz,
having a permittivity of 120, what will the Coulomb force
be between them if they are 1.0 cm apart?
SOLUTION:
F = (1/[4])q1q2 / r 2
= (1/[4128.8510-12]) (1.6010-19)2 / 0.0102
= 1.9210-25 N.
Coulomb’s law – extended distribution
 Coulomb’s law works not only for point charges, which have no
radii, but for any spherical distribution of charge at any radius.
 Be very clear that r is the distance between the centers of the charges.
Q
q
r
EXAMPLE: A conducting sphere of radius 0.10 m holds an electric charge of
Q = +125 C. A charge q = -5.0 C is located 0.30 m from the surface of Q.
Find the electric force between the two charges.
 r = 0.10 + 0.30 = 0.40 m
 F= kQq / r 2
= 8.99109 2510-6 5.010-6 / 0.40 2
= 35 N, toward positive charge
Question
Say the electric field from an isolated point charge has
a certain value at a distance of 1m. How will the electric
field strength compare at a distance of 2 m from the
charge?
It will be ¼ as much – inverse square law for force
between two charges carries over to the electric field
from a point charge.
Solving problems involving electric fields
EXAMPLE: Suppose test charges are placed at points A
and B in the electric field of the dipole, as shown. Trace
their paths when released.
SOLUTION:
Just remember: Test charges travel
with the field arrows and on the field lines.
C
A
PRACTICE: Suppose small negative charges
are placed at points C and D in the electric
field of the dipole, as shown. Trace their
paths when released.
B
D
SOLUTION:
Just remember: (-) charges travel
against the field arrows and on the
field lines.
Electric field – sketching
PRACTICE: We can simplify our drawings of electric
fields by using top views and using rays. Which field is
that of the…
(a) Largest negative charge?
D
(b) Largest positive charge?
A
(c) Smallest negative charge?
C
(d) Smallest positive charge?
E
SOLUTION: The larger the charge, the more
concentrated the field.
Lines show the direction a positive test charge will
go.
Outward is (+) charge, inward (-).
A
B
D
C
E
F
Question?
What is the direction of the electric field at point C?
1) Left
2) Right
3)
Zero
Away from positive charge (right)
Towards negative charge (right)
y
Net E field is to right.
C
x
Question?
What is the direction of the electric field at point A?
1) Up
2) Down
3) Left
4) Right
5) Zero
A
x
Question?
What is the direction of the electric field at point B?
1) Up
2) Down
3) Left
4) Right
5) Zero
y
B
x
Question?
What is the direction of the electric field at point A, if the
two positive charges have equal magnitude?
1) Up
2) Down
3) Left
4) Right
5) Zero
A
x
Solving problems involving electric fields
𝐸=𝑘
EXAMPLE: Two charges of -0.225 C each are located at
opposite corners of a square having a side
length of 645 m. Find the electric field vector at
𝑄
𝑟2
E2
q1
(a) the center of the square, and
(b) one of the unoccupied corners.
s
E1
(a)
SOLUTION: Start by making a sketch.
(b) The two fields are at right angles.
E1 =
(8.99109)(0.225)
2
/ 645 =
().
4860 NC-1
E2 = (8.99109)(0.225) / 645 2 = 4860 NC-1
q2
s
(a)The opposing fields cancel so E = 0.
().
E2 = E12 + E22 = 2(4860)2 = 47239200 E = 6870 NC-1.
(b)
E1
E2
sum points to
center of
square
Solving problems involving electric fields
PRACTICE:
Two stationary charges are shown. At which point is the electric
field strength the greatest?
SOLUTION:
Sketch in the field due to each charge at each point.
Fields diminish as 1 / r2.
Fields point away from (+) and toward (-).
The only place the fields add is point B.
Coulomb’s law – extended distribution
 Coulomb’s law works not only for point charges, which have no
radii, but for any spherical distribution of charge at any radius.
 Be very clear that r is the distance between the centers of the charges.
Q
q
r
EXAMPLE: A conducting sphere of radius 0.10 m holds an electric charge of
Q = +125 C. A charge q = -5.0 C is located 0.30 m from the surface of Q.
Find the electric force between the two charges.
 r = 0.10 + 0.30 = 0.40 m
 F= kQq / r 2
= 8.99109 2510-6 5.010-6 / 0.40 2
= 35 N, toward positive charge
Solving problems involving electric fields
PRACTICE: An isolated metal sphere of radius 1.5 cm has a charge of -15 nC
placed on it.
(a) Sketch in the electric field lines outside the sphere.
(b) Find the electric field strength at the surface of the sphere.
(c) An electron is placed on the outside surface of the sphere and released.
What is its initial acceleration?
SOLUTION:
(a) Field lines point towards (-) charge.
(b) The field equation works as if all of the charge is
(-)
at the center of the spherical distribution.
E = kQ / r 2
9)(1510
2 = 6.0×105 NC-1.
= (8.9910
0.015
(c) The electron
feels
force F-9=) /Eq
so that
F = Ee = (6.0105)(1.610-19) = 9.610-14 N.
a=F/m
= (9.610-14) / (9.1110-31) = 1.11017 m s-2.
Solving problems involving electric fields
PRACTICE: If the charge on a 25 cm radius metal sphere is +150 C,
calculate
(a) the electric field strength at the surface.
(b) the field strength 25 cm from the surface.
(c) the force on a -0.75 C charge placed 25 cm from the surface.
SOLUTION: Use E = kQ / r2, and for (c) use E = F / q.
(a) E = kQ / r2
= (8.99109)(15010-6) / 0.25 2 = 2.2107 NC-1.
(b) E = (8.99109)(15010-6) / 0.50 2 = 5.4106 NC-1.
(c) F = Eq = (5.4106)(-0.7510-6) = -4.0 N.
The minus sign means it is an attractive force.
PRACTICE: The uniform electric field strength inside the parallel plates is
275 N C-1. A +12 C charge having a mass of 0.25 grams is placed in the
field at A and released.
(a) What is the electric force acting on the charge?
(b) What is the weight of the charge?
SOLUTION:
(a) F = Eq = (275)(1210-6) = 0.0033 N.
E
(b) F = mg = (0.00025)(9.8) = 0.0025 N.
A
(c) What is the acceleration of the charge?
SOLUTION: Use Fnet = ma.
The electric force is trying to make the charge go up, and the weight is
trying to make it go down. Thus
Fnet = 0.0033 - 0.0025 = 0.0008 N.
Fnet = ma
0.0008
= 0.00025a  a = 3.2 m s-2 ( ).
0.025 m
Electric field – between parallel plates
Potential difference
PRACTICE: A charge of q = +15.0 C is moved from point A, having
a voltage (potential) of 25.0 V to point B, having a voltage
(potential) of 18.0 V.
(a) What is the potential difference undergone by the charge?
(b) What is the work done in moving the charge from A to B?
SOLUTION:
A
B
q
(a) V = VB – VA = 18.0 – 25.0 = -7.0 V.
(b) W = qV = 15.010-6-7.0 = -1.1 10-4 J.
 Many books use V instead of V.
Potential difference – the electronvolt (1eV)
PRACTICE: An electron is moved from Point A, having a voltage
(potential) of 25.0 V, to Point B, having a voltage (potential) of 18.0 V.
(a) What is the work done (in eV and in J) on the electron
by the external force during the displacement?
A
B
q
SOLUTION:
 W = q(VB – VA). W = -e(18.0 V – 25.0 V) = 7.0 eV.
7.0 eV (1.6010-19 J / eV) = 1.12 10-18 J.
(b) If the electron is released from Point B, what is its speed when it
reaches Point A?
A
SOLUTION: EK + EP = 0 with EP = – 1.12 10-18.
EK = – EP
(1/2)mv2 – (1/2)mu2 = –(– 1.12 10-18)
(1/2)(9.1110-31)v2 = 1.12 10-18
v = 1.57106 ms-1.
q
Since the electron is more attracted to A than B, we have stored this
energy as potential energy.
B
Potential difference – path independence
EXAMPLE: A charge of q = +15.0 C is moved from point A, having a
voltage (potential) of 25.0 V to point B, having a voltage (potential)
of 18.0 V, in three different ways. What is the work done in
each case?
SOLUTION:
A
B
The work is independent of the path
because the
electric force is a conservative force.
W = qV = 15.010-6 -7.0 = -1.1 10-4 J. Same for all.
 Gravitational force is also a conservative force. You remember
that work done by gravitational force will be the same (converted
into KE) if we throw a stone from certain height with the same
speed in any direction.
Potential difference – between parallel plates
PRACTICE: Two parallel plates with plate separation d are charged
up to a potential difference of V simply by
connecting a battery (shown) to them. The electric field between
the plates is E. A positive charge q is moved from A to B.
A
(a) How much work is done in moving q
through the distance d?
(b) Find the potential difference V across the plates.
SOLUTION: W = Fd cos , F = Eq, and W = qV.
(a) W = Fd cos 0° = (Eq)d.
(b) qV = Eqd  V = Ed.
d
E
B
Potential difference – between parallel plates
PRACTICE: Two parallel plates with plate
separation 2.0 cm are charged up to the
potential difference shown. Which one of
the following shows the correct direction
and strength of the resulting electric field?
SOLUTION:
Since the greater positive is plate Y, the
electric field lines point from Y  X.
From V = Ed we see that
E =V/d
= (100 – 50) / 2
= 25 V cm-1.
Drift speed
I = nAvq
current vs. drift velocity
PRACTICE: Suppose the current in a 2.5 mm diameter
copper wire is 1.5 A and the number density of the free
electrons is 5.01026 m-3. Find the drift velocity.
SOLUTION: Use I = nAvq, where A = d 2/ 4.
A = d 2/ 4 = (2.510-3) 2/ 4 = 4.9110-6 m2.
v = I / [nAq]
= 1.5 / [5.010264.9110-61.610-19 ] = 0.0038 ms-1.
Resistance
Note that resistance depends on temperature. The IBO does not require
us to explore this facet of resistivity.
PRACTICE: What is the resistance of a 0.00200 meter long carbon
core resistor having a core diameter of 0.000100 m? Assume
the temperature is 20 C.
r = d / 2 = 0.0001 / 2 = 0.00005 m.
A = r2 = (0.00005)2 = 7.85410-9 m2.
From the table  = 360010-8  m.
R = L / A
= (360010-8)(0.002) / 7.85410-9 = 9.17 .
A
L
Examples
• If a 3 volt flashlight bulb has a resistance of 9 ohms, how
much current will it draw?
• I = V / R = 3 V / 9  = 0.33 A
• If a light bulb draws 2 A of current when connected to a
120 volt circuit, what is the resistance of the light bulb?
• R = V / I = 120 V / 2 A = 60 
Example
A copper wire has a length of 1.60 m and a diameter of 1.00 mm. If the wire is
connected to a 1.5-volt battery, how much current flows through the wire?
The current can be found from Ohm's Law, V = IR. The V is the battery
voltage, so if R can be determined then the current can be calculated.
The first step, then, is to find the resistance of the wire:
L = 1.60 m.
r = 0.5 mm
 = 1.72x10-8 m, copper - books
The resistance of the wire is then:
R = L/A = (1.72x10-8 m)(1.60)/(7.9x10-7m2 ) = 3.50 
The current can now be found from Ohm's Law:
I = V / R = 1.5 / 3.5 = 0.428 A
Ohmic and Non-Ohmic behaviour
EXAMPLE: The graph shows the applied voltage V vs resulting current I
through a tungsten filament lamp.
a. Find R when I = 0.5 mA and 1.5 mA. Is this filament ohmic or non-ohmic?
 At 0.5 mA: V = 0.08 V
R = V / I = 0.08 / 0.510-3 = 160 .
 At 1.5 mA: V = 0.6 V
R = V / I = 0.6 / 1.510-3 = 400 .
b. Explain why a lamp filament
might be non-ohmic.
 tungsten is a conductor.
 Therefore, the hotter the filament the
higher R.
But the more current, the hotter a lamp
filament burns.
Thus, the bigger the I the bigger the R.
Since R is not constant
the filament is non-ohmic.
Ohmic and Non-Ohmic behaviour
EXAMPLE: The I-V characteristic
non-ohmic compoistic for a 40  ohmic component
0.0 V to 6.0 V.
SOLUTION:
”Ohmic” means V = IR and R is
graph is linear).
Thus V = I40 or I = V / 40.
If V = 0, I = 0 / 40 = 0.0.
If V = 6, I = 6 / 40 = 0.15 A.
But 0.15 A = 150 mA.
is shown for a
nent. Sketch in the I-V characterin the range of
constant (and the
Power dissipation
PRACTICE:
The graph shows the V-I
characteristics of a tungsten
filament lamp.
What is its power consumption
at I = 0.5 mA and at I = 1.5 mA?
SOLUTION:
At 0.5 mA, V = 0.08 V.
P = IV = (0.510-3)(0.08) = 4.010-5 W.
At 1.5 mA, V = 0.6 V.
P = IV = (1.510-3)(0.6) = 9.010-4 W.
Resistors in Series
EXAMPLE: Three resistors
of 330  each are connected to
a 6.0 V battery in series
R1
R2
R3

(a) What is the circuit’s equivalent resistance?
(b) What is the current in the circuit?
(a) R = R1 + R2 + R3
R = 330 + 330 + 330 = 990 .
(b) I = V / R = 6 / 990 = 0.0061 A.
(c) What is the voltage on each resistor?
(c) The current I we just found is the same everywhere. Thus each
resistor has a current of I = 0.0061 A.
From Ohm’s law, each resistor has a voltage given by
V = IR = (0.0061)(330) = 2.0 V.
In series the V’s are different if the R’s are different.
Total resistance of
resistors in series is
greater than the
greatest
Total resistance of
resistors in parallel
is smaller than the
smallest
RESISTORS IN COMPOUND CIRCUITS
Now you can calculate current, potential drop and
power dissipated through each resistor
resistors in parallel
EXAMPLE: Three resistors of
330 
each are connected to a
6.0 V cell in
parallel as shown.
R2 R3
R1
(a) What is the circuit’s resistance?
(b) What is the voltage on each resistor?
SOLUTION:
(a) In parallel, 1 / R = 1 / R1 + 1 / R2 + 1 / R3 so that
1 / R = 1 / 330 + 1 / 330 + 1 / 330 = 0.00909.
Thus R = 1 / 0.00909 = 110 .
(b) The voltage on each resistor is 6.0 V, since the resistors are in
parallel. (Each resistor is clearly directly connected to the battery).
(c) What is the current in each resistor?
SOLUTION:
(c) Using Ohm’s law (I = V / R):
I1 = V1 / R1 = 6 / 330 = 0.018 A.
I2 = V2 / R2 = 6 / 330 = 0.018 A.
I3 = V3 / R3 = 6 / 330 = 0.018 A.
In parallel the I’s are
different if the R’s are
different.

Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Circuit diagrams - voltmeters are connected in parallel
PRACTICE: Draw a schematic diagram
for this circuit:
1.06
SOLUTION:
FYI
Be sure to position the
voltmeter across the desired
resistor in parallel.
Circuit diagrams - voltmeters are connected in parallel
EXAMPLE:
A battery’s voltage is measured as shown.
(a) What is the uncertainty in it’s measurement?
SOLUTION:
For digital devices always use the place value of
the least significant digit as your raw uncertainty.
For this voltmeter the voltage is measured
to the tenths place so we give the raw
uncertainty a value of ∆V = 0.1 V.
09.4
00.0
(b) What is the fractional error in this
measurement?
SOLUTION: Fractional error is just V / V.
For this particular measurement :
V / V = 0.1 / 9.4 = 0.011 (or 1.1%).
When using a voltmeter the red lead is placed at the point of highest potential.
Potential divider circuits
PRACTICE:
Find the output voltage if the battery has
an emf of 9.0 V, R1 is a 2200  resistor,
and R2 is a 330  resistor.
SOLUTION:
 VOUT = VIN [ R2 / (R1 + R2) ]
VOUT = 9 [ 330 / (2200 + 330) ]
VOUT = 9 [ 330 / 2530 ] = 1.2 V.
PRACTICE:
Find the value of R2 if the battery has an emf of 9.0 V, R1 is a 2200  resistor,
and we want an output voltage of 6 V.
SOLUTION:
 VOUT = VIN [ R2 / (R1 + R2) ] 
6 = 9 [ R2 / (2200 + R2) ]
6(2200 + R2) = 9R2  13200 = 3R2
R2 = 4400 
The bigger R2 is in comparison to R1, the closer VOUT
is in proportion to the total voltage.
Potential divider circuits
PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W”
on its package. The potentiometer has a resistance
from X to Z of 24  and has linear variation.
(a) Sketch the variation of the p.d. V vs. the current I for
a typical filament lamp. Is it ohmic? ohmic means linear
SOLUTION: Since the temperature increases with
the current, so does the resistance.
But from V = IR we see that R = V / I, which is the slope.
Thus the slope should increase with I.
R1
R2
(b) The potentiometer is adjusted so that the meter shows
4.0 V. Will it’s contact be above Y, below Y, or exactly on Y?
SOLUTION: The circuit is acting like a potential
divider with R1 being the resistance between X and Y
and R2 being the resistance between Y and Z.
V
non-ohmic
Since we need VOUT = 4 V, and since VIN = 6 V,
the contact must be adjusted above the Y.
I
PRACTICE: A light sensor consists of a 6.0 V battery, a 1800 Ω resistor and a light-dependent
resistor in series. When the LDR is in darkness the pd across the resistor is 1.2 V.
(a) Calculate the resistance of the LDR when it is in darkness.
(b) When the sensor is in the light, its resistance falls to 2400 Ω. Calculate the pd across the
LDR.
(a) As the pd across the resistor is 1.2 V, the pd across the
LDR must be 6-1.2=4.8.
The current in the circuit is
The resistance of the LDR is
(b)
For the ratio of pds to be 1.33, the pds must be
2.6 V and 3.4 V with the 3.4 V across the LDR.
Potential divider circuits
PRACTICE: A light-dependent resistor (LDR) has R = 25  in
bright light and R = 22000  in low light.
An electronic switch will turn on a light when its p.d. is above
7.0 V. What should the value of R1 be?
SOLUTION: VOUT = VIN [ R2 / (R1 + R2) ]
7
= 9 [ 22000 / (R1 + 22000) ]
7(R1 + 22000) = 9(22000)
7R1 + 154000 = 198000
R1 = 6300  (6286)
Potential divider circuits
PRACTICE: A thermistor has a resistance of 250  when it
is in the heat of a fire and a resistance of 65000  when
at room temperature.
An electronic switch will turn on a sprinkler system
when its p.d. is above 7.0 V.
(a) Should the thermistor be R1 or R2?
SOLUTION:
Because we want a high voltage at a high
temperature, and because the thermistor’s
resistance decreases with temperature,
it should be placed at the R1 position.
(b) What should R2 be?
SOLUTION: In fire the thermistor is R1 = 250 .
7
= 9 [ R2 / (250 + R2) ]
7(250 + R2) = 9R2
R2 = 880  (875)
Potential divider circuits
PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W”
on its package. The potentiometer has a resistance
from X to Z of 24  and has linear variation.
(a) Sketch the variation of the p.d. V vs. the current I for
a typical filament lamp. Is it ohmic? ohmic means linear
SOLUTION: Since the temperature increases with
the current, so does the resistance.
But from V = IR we see that R = V / I, which is the slope.
Thus the slope should increase with I.
R1
R2
(b) The potentiometer is adjusted so that the meter shows
4.0 V. Will it’s contact be above Y, below Y, or exactly on Y?
SOLUTION: The circuit is acting like a potential
divider with R1 being the resistance between X and Y
and R2 being the resistance between Y and Z.
V
non-ohmic
Since we need VOUT = 4 V, and since VIN = 6 V,
the contact must be adjusted above the Y.
I
PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W”
on its package. The potentiometer has a resistance
from X to Z of 24  and has linear variation.
(c) The potentiometer is adjusted so that the meter shows 4.0 V.
What are the current and the resistance of the lamp at this instant?
SOLUTION: P = 0.80 W and V = 4.0 V.
 P = IV  0.8 = I(4)  I = 0.20 A.
 V = IR  4 = 0.2R  R = 20. .
You could also use P = I 2R for this last one.
(d) The potentiometer is adjusted so that the meter shows 4.0 V.
What is the resistance of the Y-Z portion of the potentiometer?
SOLUTION: Let R1 = X to Y and R2 = Y to Z resistance.
Then R1 + R2 = 24 so that R1 = 24 – R2.
From VOUT = VIN [ R2 / (R1 + R2) ] we get
4 = 7 [ R2 / (24 – R2 + R2) ]  R2 = 14  (13.71).
R1
R2
PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W”
on its package. The potentiometer has a resistance
from X to Z of 24  and has linear variation.
(e) The potentiometer is adjusted so that the meter shows 4.0
V. What is the current in the Y-Z portion of the potentiometer?
SOLUTION:
V2 = 4.0 V because it is in parallel with the lamp.
I2
= V2 / R2
= 4 / 13.71 = 0.29 A
(f) The potentiometer is adjusted so that the meter shows 4.0 V. What is
the current in the ammeter?
SOLUTION: The battery supplies two currents.
The red current is 0.29 A because it is the I2 we just calculated in (e).
The green current is 0.20 A found in (c).
The ammeter has both so I = 0.29 + 0.20 = 0.49 A.
R1
R2
Solving problems involving circuits
PRACTICE: A battery is connected to a 25-W lamp as shown.
What is the lamp’s resistance?
SOLUTION:
Suppose we connect
a voltmeter to the
circuit.
We know P = 25 W.
We know V = 1.4 V.
From P = V 2 / R we get
R = V 2/ P = 1.4 2 / 25
= 0.078 .
01.4
00.0
Solving problems involving circuits
PRACTICE: Which circuit shows the correct
setup to find the V-I characteristics of a
filament lamp?
SOLUTION:
The voltmeter must be in
parallel with the lamp.
It IS, in ALL cases.
The ammeter must be in series
with the lamp and must read
only the lamp’s current.
two
currents
lamp
current
no
currents
short
circuit!
Solving problems involving circuits
PRACTICE: A non-ideal voltmeter is used to measure
the p.d. of the 20 k resistor as shown. What will
its reading be?
SOLUTION: There are two currents in the circuit
because the voltmeter does not have a high enough
resistance to prevent the green one from flowing.
The 20 k resistor is in parallel with the 20 k:
1 / R = 1 / 20000 + 1 / 20000 = 2 / 20000.
R = 20000 / 2 = 10 k.
But then we have two 10 k resistors in series and
each takes half the battery voltage, or 3 V.
equivalent ckt
Solving problems involving circuits
PRACTICE: All three circuits use the same resistors and the same cells.
Highest I
0.5R
parallel
Lowest I
2R
series
Middle I
1.5R
R
0.5R
combo
Which one of the following shows the correct ranking for the currents
passing through the cells?
SOLUTION: The bigger the R the smaller the I.
Kirchhoff’s second law
in a complete circuit loop, the sum of the emfs in the loop is
equal to the sum of the potential differences in the loop or
the sum of all variations of potential in a closed loop equals
zero
𝑓𝑜𝑟 𝑎𝑛𝑦 𝑐𝑙𝑜𝑠𝑒𝑑 𝑙𝑜𝑜𝑝 𝑖𝑛 𝑎 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ∑𝜀 = ∑𝐼𝑅
Equivalent to conservation of energy.
Loop GABCDEG travelling
anticlockwise round the loop
The direction of loop travel and the current direction are in all cases the
same. We give a positive sign to the currents when this is the case. The emf
of the cell is driving in the same direction as the loop travel direction; it gets
a positive sign as well. If the loop direction and the current or emf were to
be opposed then they would be given a negative sign.
𝜀 = 𝐼1 𝑅1 + 𝐼2 𝑅2
loop EFGE travelling
clockwise round the loop
resistor R3: loop direction is in the same direction as the conventional current.
resistor R1: loop direction is in opposite direction to the conventional current,
so there has to be a negative sign.
There is no source of emf in the loop so the Kirchhoff equation becomes
0 = 𝐼3 𝑅3 − 𝐼2 𝑅2
Kirchhoff’s first law can be applied at point G.
The total current into point G is
𝜀 = 𝐼1 𝑅1 + 𝐼2 𝑅2
0 = 𝐼3 𝑅3 − 𝐼2 𝑅2
𝐼1 = 𝐼2 + 𝐼3
Kirchhoff’s rules – solving the circuit
PRACTICE: Calculate the unknown branch current in the following
junctions.
I = 9 + 5 – 3 = 11 A out of the junction.
I = 12 – 14 = –2 A directed
away from the junction.
Kirchhoff’s rules – solving the circuit
PRACTICE: Calculate the currents in the circuit shown.
For loop 1
3 = 3I1 + 9I2
1]
[equation
Current directions have been assigned and two
loops 1 and 2 and junction A defined in the diagram.
(the emf in the loop is 3 V)
For loop 2
0 = 6I3 – 9I2 [equation 2]
(there is no source of emf in this loop, current I2 is
in the opposite direction to the loop direction 0.
For junction A
I1 = I2 + I3
3 equations with three
unknowns:
I1 = 0.45 A
I2 = 0.18 A
I3 = 0.27 A
Kirchhoff’s rules – solving the circuit
EXAMPLE: Suppose each of the resistors is R = 2.0 , and the emfs are
1 = 12 V and 2 = 6.0 V. Find the voltages and the currents of the circuit.
▪ I1 – I2 + I3 = 0 (1)
+ –V2 + –V4 + 1 + – 2 = 0,
– – V – V = 0
2
3
4
▪ –V
▪
–V
1
1
–
2
+ –V2 + –V4 + 1 + – 2 = 0 & V=IR  –2I1 + –2I1 + –2I2 + 12 + –6 = 0 (2)
– V3 – V4 = 0 & V=IR 
–– 6
We now have three equations in I:
(1)  I3 = I2 – I1.
I1 = 1.8 A
(2)  3 = 2I1 + I2.
I2 = -0.6 A
I3 = -2.4 A
(3)  3 = -I2 + -I3
– 2I3 – 2I2 = 0 (3)
Finally, we can redraw our currents:
resistor voltages: V = IR
V1 = 1.8(2) = 3.6 V.
V2 = 1.8(2) = 3.6 V.
V3 = 2.4(2) = 4.8 V.
V4 = 0.6(2) = 1.2 V
Topic 5: Electricity and magnetism
5.3 – Electric cells
Cells – electric potential difference
EXAMPLE:
200 C of charge is brought from an electric potential of
2.0 V to an electric potential of 14 V through use of a
car battery. What is the change in potential energy of
the charge?
SOLUTION:
From ∆V = ∆EP / q we see that ∆EP = q∆V. Thus
∆EP = q(V – V0)
∆EP = (20010-6)(14 – 2)
∆EP = 0.0024 J.
FYI
Note the increase rather than the decrease.
Topic 5: Electricity and magnetism
5.3 – Electric cells
Cells – electromotive force (emf)
EXAMPLE: How much chemical energy is converted to
electrical energy by the cell if a charge of 15 C is
drawn by the voltmeter?
SOLUTION:
01.6
From ∆V = ∆EP / q
we have  = ∆EP / q.
Thus
∆EP = q
= (1.6)(1510-6)
= 2.410-5 J.
Topic 5: Electricity and magnetism
5.3 – Electric cells
Cells – internal resistance
PRACTICE: A battery has an internal resistance of r =
1.25 . What is its rate of heat production if it is
supplying an external circuit with a current of I = 2.00 A?
SOLUTION:
Rate of heat production is power.
P = I 2r
P = (2 2)(1.25) = 5.00 J s-1 (5.00 W.)
FYI
If you double the current, the rate of heat generation
will quadruple because of the I 2 dependency.
If you accidentally “short circuit” a battery, the battery
may even heat up enough to leak or explode!
Topic 5: Electricity and magnetism
5.3 – Electric cells
1.6 V
1.5 V
Cells – internal resistance
EXAMPLE:
A cell has an unloaded voltage of 1.6 V
and a loaded voltage of 1.5 V when a
330  resistor is connected as shown.
(a) Find .
(b) Find I.
(c) Find the cell’s internal resistance.
SOLUTION:
(a) From the first schematic we see
that  = 1.6 V. (Unloaded cell.)
(b) From the second diagram we see that the voltage
across the 330  resistor is 1.5 V. (Loaded cell.)
V = IR so that 1.5 = I(330) and I = 0.0045 A.
Topic 5: Electricity and magnetism
5.3 – Electric cells
1.6 V
1.5 V
Cells – internal resistance
EXAMPLE:
A cell has an unloaded voltage of 1.6 V
and a loaded voltage of 1.5 V when a
330  resistor is connected as shown.
(a) Find .
(b) Find I.
(c) Find the cell’s internal resistance.
SOLUTION:
(c) Use the emf relationship  = IR + Ir.
1.6 = (0.0045)(330) + (0.0045)r
1.6 = 1.5 + 0.0045r
0.1 = 0.0045r
 r = 22 .
Topic 5: Electricity and magnetism
5.3 – Electric cells
1.6 V
1.5 V
Cells – terminal potential difference
EXAMPLE:
A cell has an unloaded voltage of 1.6 V
and a loaded voltage of 1.5 V when a
330  resistor is connected as shown.
(d) What is the terminal potential
difference t.b.d. of the cell?
SOLUTION:
(d) The terminal potential difference is
the potential difference at the terminals
of the cell where it connects to the external circuit.
This cell has an unloaded t.p.d of 1.6 V, and a loaded
t.p.d. of 1.5 V. Note that the t.p.d. depends on the load.
Solving problems involving magnetic fields
PRACTICE: Find the magnetic flux density 1.0 cm from a
straight wire carrying a current of 25 A.
SOLUTION: Magnetic flux density is just B.
Use B = 0I / (2d) where d = 1.0 cm = 0.010 m.
B = 410-725 / [20.010] = 5.010-4 T.
PRACTICE: Find the B-field strength at the center of a 1.0 cm
radius loop of wire carrying a current of 25 A.
SOLUTION:
Use B = 0I / (2R) where R = 1.0 cm = 0.010 m.
B = 410-7times
25 / [20.010]
stronger! = 1.610-3 T.
RHR for solenoids
Although you can use simply RHR 2 to determine direction of magnetic
field inside solenoid, there is simple RHR for solenoids:
Grasp the solenoid with your right hand in such a way that your fingers
curl in the direction of the current.
Your extended thumb points in the direction of north pole.
PRACTICE: In the solenoid shown
label the north and south poles.
I
I
PRACTICE: The north and south
poles are labeled in the solenoid.
Sketch in the current, both
entering and leaving the solenoid.
Solving problems involving magnetic fields and forces
x x x x x x x INTO Page
•••••••••••••
OUT of Page
Right Hand Rule for
mag force
practice
A proton enters a magnetic
field, as shown. Which way
will the electron turn?
1.
2.
3.
4.
Up
Down
Out of page
Into page
Put your fingers in
the direction of the
velocity and curl out
of the page … your
thumb points up
Solving problems involving magnetic fields and forces
x x x x x x x INTO Page
•••••••••••••
OUT of Page
Right Hand Rule for
mag force
practice
An electron enters a magnetic
field, as shown. Which way
will the electron turn?
1.
2.
3.
4.
Up
Down
Out of page
Into page
Remember to flip
the direction of the
force for negative
charges
Solving problems involving magnetic fields and forces
In which direction will wire segment B be pushed?
1.
2.
3.
4.
5.
Up
Down
Out of page
Into page
No force exists
Right Hand Rule for
mag force
practice
In which direction will wire segment C be pushed?
1.
2.
3.
4.
5.
Up
Down
Out of page
Into page
No force exists
V and B are in the
same direction;
no force exists.
Solving problems involving magnetic fields and forces
EXAMPLE: The tendency of a moving
charge to follow a curved trajectory in
a magnetic field is used in a
mass spectrometer
An unknown element is ionized,
and accelerated by an applied
voltage in the chamber S.
It strikes a phosphorescent
screen and flashes.
By measuring x, one can determine the
mass of the ion, and hence the unknown
Show that m = xqB / 2v
SOLUTION:
 r = mv / qB.
 m = rqB / v.
 from the picture we see that r = x / 2.
 m = rqB / v = xqB / 2v.
Force on a current-carrying conductor in a B-field
PRACTICE: A piece of aluminum foil is held between the
two poles of a strong magnet as shown.
When a current passes through the foil in the direction
shown, which way will the foil be deflected?
SOLUTION:
F
Sketch in B and v:
Use the RHR for a moving
charge
in a B-field.
B
v
Question
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1
with velocity v1= 75 m/s up, and follows the
dashed trajectory.
2
1
v = 75 m/s
q = +25 mC
What is the speed of the particle in chamber 2?
1) v2 < v1
2) v2 = v1
3) v2 > v1
Magnetic force is always perpendicular to velocity, so it changes direction,
not speed of particle.
43
Question
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1
with velocity 75 m/s up, and follows the
dashed trajectory.
2
1
v = 75 m/s
q = +25 mC
Compare the magnitude of the magnetic field in
chambers 1 and 2
1) B1 > B2
2) B1 = B2.
3) B1 < B2
Larger B, greater force, smaller R
mv
R
qB
Question
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1
with velocity 75 m/s up, and follows the
dashed trajectory.
2
1
v = 75 m/s
q = ?? mC
mv
R
qB
A second particle with mass 2m enters the chamber
and follows the same path as the particle with mass
m and charge q=25 mC. What is its charge?
1) Q = 12.5 mC
2) Q = 25 mC
3) Q = 50 mC
If both mass and charge double there is no
change in R
Solving problems involving magnetic fields and forces
PRACTICE: A 25 C charge traveling at 150 m s-1 to the north enters
a uniform B-field having a strength of 0.050 T and pointing to the west
(a) What will be the magnitude of the
magnetic force acting on the charge?
(b) Which way will the charge be deflected?
UP
F
SOLUTION: sketch would be helpful:
(a) F = qvB sin  = (2510-6)(150)(0.050) sin 90°
F = 1.910-4 N.
(b) Use the RHR for force on q in mag. field.
q will deflect upward.
(c) Explain why the magnetic force can not
change the magnitude of the velocity of the
charge while it is being deflected.
Since 𝐹 is perpendicular to 𝑣 it acts as centripetal force:
only direction of 𝑣 will change, not its magnitude
W
B
q
S
DOWN
N
v
E
(d) How do you know that the charge will be in uniform circular motion?
SOLUTION:
 v is constant.
 Since q and v and B are constant, so is F.
 Since F is constant, so is a.
 A constant acceleration perpendicular to the charge’s velocity is the
definition of centripetal acceleration causing uniform circular motion.
(e) If the charge has a mass of 2.510-5 kg, what will be the radius of
its circular motion?
SOLUTION:
In (a) we found that F = 1.910-4 N.
Then a = F / m = 1.910-4 / 2.510-5 = 7.6 m s-2.
From (d) we know the charge is in UCM.
Thus a = v2 / r so that
r = v2 / a = 1502 / 7.6 = 3000 m. (2961)
Magnetic Field & Magnetic Force Problems
We do:
What is the minimum magnetic field necessary to exert a 5.4
X 10-15 N force on an electron moving at 2.1 X 107 m/s?
B = F / qvsinθ
B will be at a minimum when sin θ = 1
B = F / qv = 5.4X10-15N / (1.6 X 10-19 C X 2.1 X 107 m/s)
B = 1.61 X 10-3 T
Magnetic Field & Magnetic Force Problems
You do:
What is the magnetic field necessary to exert a 5.4 X 10-15 N
force on an electron moving at 2.1 X 107 m/s if the magnetic
field is at 45 degrees from the electron’s velocity?
B = F / qvsinθ = 5.4X10-15N / (1.6 X 10-19 C X 2.1 X 107 m/s X
sin 45)
B = 2.3 X 10-3 T.
Magnetic Field & Magnetic Force Problems
We do and You do
What is the magnitude of the magnetic force on a proton
moving at 2.5 X 105 m/s in a magnetic field of 0.5 T …
(a) …if the velocity and magnetic field are at right angles?
(b) … if the velocity and magnetic field are at 30°?
(c) … if the velocity is parallel to a magnetic field?
F = qvBsinθ , so
(a) when θ = 90°, F = (1.6 X 10-19 C)(2.5 X 105 m/s)(0.5 T) =
2.0 X 10-14 N,
(b) F = (2.0 X 10-14 N) sin 30° = 1.0 X 10-14 N, and
(c) F = qvB sin 0° = 0.
1.
What is the direction of the
magnetic field at point A?
a)
b)
c)
d)
2.
A
What is the direction of the
magnetic field at point B?
a)
b)
c)
d)
3.
Into the page
Out of the page
Up
Down
I
B
Into the page
Out of the page
Up
Down
What is the shape of the
magnetic field lines?
a) circles
b) spirals
c) radially outward
4.
Where is the magnetic field stronger?
a) point A
b) point B
c) it’s the same at A and B
Force between wires carrying current
Use RHR #2 to find the direction of the
magnetic field at point P
I up
I up
F
B
F xP
I down
I up
F
B
xF
P
use RHR #1 to find the force on second wire
Conclusion: Currents in same
direction attract!
Use RHR #2 to find the direction of the
magnetic field at point P
use RHR #1 to find the force on
second wire
Conclusion: Currents in opposite
direction repel!
What is the direction of the force on the top
wire, due to the two below?
1) Left 2) Right
3) Up
4) Down 5) Zero
What is the direction of the force
on the middle wire, due to the two
others?
I
1) Left
I
2) Right
I
3) Up
4) Down 5) NoneZero
What is the direction of the force
on the left wire, due to the two
others?
I
1) Left
I
2) Right
I
3) Up
4) Down 5) None 5) Zero
What is the direction of the force
on the middle wire, due to the two
others?
I
1) Left
2I
2) Right
3I
3) Up
4) Down 5) None 5)
Zero
What is the direction of the force
on the middle wire, due to the two
others?
I
1) Left
I
2) Right
I
3) Up
4) Down 5) None5) Zero
What is the direction of the
magnetic field on a point P in the
middle of two wires?
I
I
P
What is the direction of the force
on the left, due to the two others?
I
I
I
X
1) Left
2) Right
3) Up
4) Down 5) None) Zero
1) Left
2) Right
3) Up
4) Down 5) None 5) Zero
Solving problems involving magnetic fields and forces
PRACTICE: Two parallel wires are shown with the
given currents in the given directions. The force on
Wire 2 due to the current in Wire 1 is F. Find the force
in Wire 1 due to the current in Wire 2 in terms of F.
SOLUTION:
Recall Newton’s 3rd law.
The force on Wire 1 and the force on Wire 2
are an action-reaction pair.
But action-reaction pairs have equal magnitude
(and opposite direction).
Thus Wire 1 feels the exact same force F!
Solving problems involving magnetic fields and forces
PRACTICE: A very flexible wire
is formed into exactly two loops.
The top loop is firmly anchored
Ito a support, and cannot move.
Explain why, when a current is passed
through the wire, the
loops get closer together.
I
SOLUTION: Two parallel currents attract each other
F
I
F