Download Electric Circuits and Electromagnetism Review

ELECTRIC CIRCUITS
Conventional current is the flow of positive charges though a closed circuit. The current
through a resistance and the voltage which produces it are related by Ohm’s law. Power
is the rate at which energy is consumed in a circuit through the resistors. Resistors in a
circuit may be connected in series or in parallel. If a capacitor is placed in a circuit with
a resistor, the current in the circuit becomes time-dependent as the capacitor charges or
discharges.
QUICK REFERENCE
Important Terms
ammeter
device used to measure electrical current
ampere
unit of electrical current equal to one coulomb per second
battery
device that converts chemical energy into electrical energy , creating a potential
difference (voltage)
capacitive time constant
the product of the resistance and the capacitance in a circuit; at a time equal to the
product RC, the capacitor has reached 63% of its maximum charge
capacitor
two oppositely charged conductors used to store charge and energy in an electric
field between them
direct current
electric current whose flow of charges is in one direction only
electric circuit
a continuous closed path in which electric charges can flow
electric current
flow of charged particles; conventionally, the flow of positive charges
electric power
the rate at which work is done or energy is dissipated through a resistor
electrical resistance
the ratio of the voltage across a device to the current running through it
electron flow
the movement of electrons through a conductor; electron flow is equal and
opposite to conventional current flow
emf
electromotive force; another name for voltage
equivalent resistance
the single resistance that could replace the individual resistances
in a circuit and produce the same result
ohm
the SI unit for resistance equal to one volt per ampere
1
Ohm’s law
the ratio of voltage to current in a circuit is a constant called resistance
parallel circuit
an electric circuit which has two or more paths for the current to follow,
allowing each branch to function independently of the others
resistivity
the constant which relates the resistance of a resistor to its length and crosssectional area
resistor
device designed to have a specific resistance
schematic diagram
a diagram using special symbols to represent a circuit
series circuit
an electric circuit in which devices are arranged so that charge flows
through each equally.
terminal voltage
the actual voltage across the positive and negative terminals of a battery, which is
usually lowered by the internal resistance of the battery
voltage
the potential difference between the positive and negative sides of a circuit
element
watt
the SI unit for power equal to one joule of energy per second
2
Equations and Symbols
q
t
V
R
I
L
R
A
P  IV
I
where
I = current
Δq = amount of charge passing a given
point
Δt = time interval
R = resistance
V = voltage
P = power
RS = total resistance in series
RP = total resistance in parallel
CP = total capacitance in parallel
CS = total capacitance in series
q = charge
UE = electrical energy stored in a
capacitor
τ = capacitive time constant
Rs  R1  R2  R3  ...
1
1
1
1



 ...
R p R1 R2 R3
C P  C1  C 2  C 3  ...
1
1
1
1



 ...
C s C1 C 2 C 3
q  CV
1
1
1 Q2
CV 2  QV 
2
2
2 C
  RC
UE 
Electromotive Force and Current, and Ohm’s Law
When we connect a battery, wires, and a light bulb in a circuit shown, the bulb lights up.
But what is actually happening in the circuit?
R
I
I
V
+
-
+
-
Recall from an earlier chapter that the battery has a potential difference, or voltage,
across its ends. One end of the battery is positive, and the other end is negative. We say
that the movement of positive charge from the positive end of the battery through the
circuit to the negative end of the battery is called conventional current, or simply current.
Current is the amount of charge moving through a cross-sectional area of a conductor per
second, and the unit for current is the coulomb/second, or ampere. We use the symbol I
for current.
Area
I
+q
3
q
t
Conventional current is defined as the flow of positive charge on the AP Physics B exam,
although electrons are actually moving in the wire.
In a circuit such as the one shown above, the current is directly proportional to the
voltage. The ratio of voltage to current is a constant defined as the resistance, and is
measured in ohms (), or volts/amp. The relationship between voltage, current, and
resistance is called Ohm’s law:
R
V
I
This relationship typically holds true for the purposes of the AP Physics B exam.
Resistance and Resistivity
As charge moves through the circuit, it encounters resistance, or opposition to the flow of
current. Resistance is the electrical equivalent of friction. In our circuit above, the wires
and the light bulb would be considered resistances, although usually the resistance of the
wires is neglected. The resistance of a resistor is proportional to the length L of the
resistor and inversely proportional to the cross-sectional area A of the resistor by the
equation
R
L
ρ
A
A
L
where the constant  is called the resistivity of the resistor and has units of ohm-meters
( m). The resistivity of a material is a characteristic of the material itself rather than a
particular sample of the material.
Example 1
A copper wire has a cross-sectional area of 5.0 x 10-7 m2 and a length of 10.0 m. An
aluminum wire of exactly the same dimensions is welded to the end of the copper wire.
the ends of this long copper-aluminum wire are connected to a 3.0-volt battery. Neglect
the resistance of any other wires in the figure.
4
Cu
Al
V
Figure not drawn to scale
Determine
(a) the total resistance of the circuit.
(b) the total current in the wire.
Solution
(a) The total resistance is equal to the sum of the copper and aluminum resistors. We
can find the value of the resistivity of copper and aluminum in the table in this chapter of
the textbook.
 L 1.72 x10 8 m 10 m 
RCu  Cu 
 0.34 
A
5 x10 7 m 2
 L 2.82 x10 8 m 10 m 
R Al  Al 
 0.56 
A
5 x10 7 m 2
Rtotal  RCu  R Al  0.34   0.56   0.90 




(b) According to Ohm’s law,
3.0 V
V
I

 3.33 A
Rtotal 0.90 
Electric Power
In an earlier chapter we defined power as the rate at which work is done, or the rate at
which energy is transferred. When current flows through a resistor, heat is produced, and
the amount of heat produced in joules per second is equal to the power in the resistor. The
heating in the resistor is called joule heating.The equation that relates power to the
current, voltage, and resistance in a circuit is
P  IV  I 2 R 
V2
R
The unit for power is the joule/second, or watt.
Example 2
A simple circuit consists of a 12-volt battery and a 6  resistor. Let’s draw a schematic
diagram of the circuit, and include an ammeter which measures the current through the
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resistor, and a voltmeter which measures the voltage across the resistor, indicate the
reading on the ammeter and voltmeter, and find the power dissipated in the resistor.
V
R=6Ω
I
A
+
12V
-
Note that the ammeter is placed in line (series) with the resistor, and the voltmeter is
placed around (parallel) the resistor.
The ammeter reads the current, which we can calculate using Ohm’s law:
V 12V
I 
 2A
R 6
The voltmeter will read 12 V, since the potential difference across the resistor must be
equal to the potential difference across the battery. As we will see later, if there were
more than one resistor in the circuit, there would not necessarily be 12 volts across each.
The power can be found by
P = IV = (2 A)(12 V) = 24 watts
Series Wiring
Two or more resistors of any value placed in a circuit in such a way that the same current
passes through each of them is called a series circuit. Consider the series circuit below
which includes a voltage source 
emf, an older term for voltage) and
three resistors R1, R2, and R3.
I1
R1
ε
I2
R2
R3
6
I3
The rules for dealing with series circuits are as follows:
1. The total resistance in a series circuit is the sum of the individual resistances:
Rtotal = R1 + R2 + R3
2. The total current in the circuit is
V
I total  total .
Rtotal
This current must pass through each of the resistors, so each resistor also gets Itotal, that is,
Itotal = I1 = I2 = I3.
3. The voltage divides proportionally among the resistances according to Ohm’s law:
V1  I1 R1 ; V2  I 2 R2 ; V3  I 3 R3 ;
Example 3
Consider three resistors of 2 , 6 , and 10  connected in series with a 9 volt battery.
Draw a schematic diagram of the circuit which includes an ammeter to measure the
current through the 2 
er to measure the voltage across the 10 
resistor, and indicate the reading on the ammeter and voltmeter.
I1
A
2Ω
ε=9V
6Ω
10Ω
I3
V
7
I2
The current through each resistor is equal to the total current in the circuit, so the
ammeter will read the total current regardless of where it is placed, as long as it is placed
in series with the resistances.
V
9V
I total  total 
 0.5 A
Rtotal 2  6  10
The voltmeter will read the voltage across the 10  resistor, which is NOT 9 volts. The 9
volts provided by the battery is divided proportionally among the resistances. The voltage
across the 10  resistor is
V10 = I10R10 = (0.5 A)(10 ) = 5 V.
Parallel Wiring
Two or more resistors of any value placed in a circuit in such a way that each resistor has
the same potential difference across it is called a parallel circuit. Consider the parallel
circuit below which includes a voltage source  and three resistors R1, R2, and R3.
ε
R1
R2
R3
The rules for dealing with parallel circuits are as follows:
1. The total resistance in a parallel circuit is given by the equation
1
Rtotal

1
1
1


R1 R2 R3
2. The voltage across each resistance is the same:
Vtotal = V1 = V2 = V3
3. The current divides in an inverse proportion to the resistance:
V
V
V
I1  1 ; I 2  2 ; I 3  3
R1
R2
R3
where V1, V2, and V3 are equal to each other.
Example 4
Three resistors of resistance 2 , 3 , and 12  are connected in parallel to a battery of
voltage 24 V.
8
Draw a schematic diagram of the circuit which includes an ammeter to measure only the
current through the 2  resistor, a voltmeter to measure the voltage across the 12 
ammeter and voltmeter.
Itot
2Ω
24 V
3Ω
I1
I2
12Ω
I1
V
A
The total resistance in the circuit is found by
1
Rtotal

1
1
1
1
1
1
6
4
1
11









R1 R2 R3 2 3 12 12 12 12 12
Notice that this fraction is not the total resistance, but
this circuit must be
1
Rtotal
. Thus, the total resistance in
12
.
11
Since the ammeter is placed in series only with the 2 
tance, it will measure the
current passing only through the 2 resistance. Recognizing that the voltage is the same
(24 V) across each resistance, we have that
I2 
V2 24V

 12 A
R2
2
Each resistance is connected across the 24 V battery, so the voltage across the 10 
resistance is 24 V.
Note that the equation for the total resistance in parallel comes from the fact that the total
current is the sum of the individual currents in the circuit, and each resistor gets the same
V
V V
voltage: total  1  2  ... , where all the V ’s are equal.
Rtotal R1 R2
9
Circuits Wired Partially in Series and Partially in Parallel
Example 5
Consider the circuit below, which is a combination of series and parallel:
R1=20Ω
R3=40Ω
ε=13V
R2=30Ω
Find
(a) the total resistance,
(b) the total current in the circuit,
(c) the voltage across each resistor, and
(d) the current through each resistor.
Solution
We see that R1 is in parallel with R2, and R3 is in series with the parallel combination of
R1 and R2.
(a) Before we can find the total resistance of the circuit, we need to find the equivalent
resistance of the parallel combination of R1 and R2:
1
1
1
1
1
3
2
5







R12 R1 R2 20 30 60 60 60
which implies that the combination of R1 and R2 has an equivalent resistance of
60

.
5
Then the total resistance of the circuit is Rtotal = R12 + R3 = 12 
 = 52 .
Vtotal 13V

 0.25 A
Rtotal 52
(c) The voltage provided by the battery is divided proportionally among the parallel
combination of R1 and R2 (with R1 and R2 having the same voltage across them), and R3.
Since R3 has the total current passing through it, we can calculate the voltage across R3:
(b) The total current in the circuit is I total 
V3 = I3R3 = (0.25 A)(40 ) = 10 V.
10
This implies that the voltage across R1 and R2 is the remainder of the 13 V provided by
the battery. Thus, the voltage across R1 and R2 is 13 V – 10 V = 3 V.
(d) The current through R3 is the total current in the circuit, 0.25 A. Since we know the
voltage and resistance of the other resistances, we can use Ohm’s law to find the current
through each.
I1 
V1
3V

 0.15 A
R1 20
I2 
V2
3V

 0.10 A
R2 30
Notice that I1 and I2 add up to the total current in the circuit, 0.25 A.
Internal Resistance and Kirchhoff’s Rules
Consider the multi-loop circuit shown below:
a
R1= 8Ω
R2= 12Ω
ε1= 6 V
R3= 20Ω
b
ε 2= 3 V
In this circuit, we have three resistors and two batteries, one on either side of junction b.
We can’t say that all the resistors are in series with each other, nor parallel, because of
the placement of the batteries. If we want to find the current in and voltage across each
resistor, we must use Kirchhoff’s rules.
Kirchhoff’s Rules for Multi-loop Circuits:
1. The total current entering a junction (like a or b in the figure above) must also leave
that junction. This is sometimes called the junction rule, and is a statement of
conservation of charge.
11
2. The sum of the potential rises and drops (voltages) around a loop of a circuit must be
zero. This is sometimes called the loop rule, and is a statement of conservation of energy.
a. If we pass a battery from negative to positive, we say that there is a rise in
potential, +.
b. If we pass a battery from positive to negative, we say that there is a drop in
potential, - .
c. If we pass a resistor against the direction of our arbitrarily chosen current, we
say there is a rise in potential across the resistor, + IR.
d. If we pass a resistor in the direction of our arbitrarily chosen current, we say
there is a drop in potential across the resistor, - IR.
Example 6
Find the current through each resistor in the multiloop circuit above.
Solution
First, let’s choose directions for the each of the currents through the resistors. If we
happen to choose the wrong direction for a particular current, the value of the current will
simply come out negative.
a
I1
R1= 8Ω
I2
R2= 12Ω
loop1
R3= 20Ω
loop2
I3
ε1= 6 V
b
ε 2= 3 V
If we apply the junction rule to junction a, then I1 and I2 are entering the junction, and I3
is exiting the junction. Then
I1 + I2 = I 3
(Equation 1)
Let’s apply the loop rule beginning at junction b. Writing the potential rises and drops by
going clockwise around loop 1 gives
+1 - I1R1 + I2R2 = 0 (Equation 2)
Note that we encountered a rise in potential (+) as we passed the battery, and a drop (I1R1) across the first resistor, and a rise (+I2R2) across the second resistor.
12
Similarly, if we write the potential rises and drops around loop 2, going clockwise
beginning at b, we get
- I2R2 – I3R3 - 2 = 0 (Equation 3)
We can solve the three equations above for the values of the three currents by substituting
the values of the resistors and the emfs of the batteries:
Equation 2:
6V  I1 8   I 2 12   0
Equation 3:
 I 2 12   I 3 20   3V  0
Solving Equations 1, 2, and 3 simultaneously gives
I1 = - 6.3 A, I2 = 9.2 A, and I3 = 2.9 A.
Note that I1 is negative, which simply means we chose the wrong direction for the current
through R1.
Capacitors in Series and Parallel
In the last chapter we discussed capacitors, two oppositely charged conductors used to
store charge. There are times when we want to know the equivalent capacitance of two or
more capacitors which are connected in a certain way. Consider two capacitors which are
connected to a battery in the figure below:
+
ε
+
+
-
-
C1
C2
-
We say that the capacitors are connected in parallel. Note that the positive plates of the
capacitors are connected to each other and the negative plates are connected to each
other. Since the positive plates are also connected to the positive terminal of the battery,
and the negative plates are connected to the negative terminal of the battery, the voltage
across each capacitor is the same as the battery voltage. In other words, capacitors in
parallel have the same voltage across them. However, the total charge will be divided
proportionally among them:
13
qtotal  q1  q 2
CtotalVtotal  C1V1  C 2V2
But the voltages are all the same :
Ctotal  C1  C 2 (parallel)
Thus, to find the total capacitance in parallel, we simply add the capacitors.
If we connect two capacitors in series as shown in the figure below, each capacitor will
get the same charge, but will divide the voltage proportionally.
+
C1
+
ε
-
+
C2
-
We can find the total (or equivalent) capacitance by summing the voltages:
Vtotal  V1  V2
qtotal q1 q 2


C total C1 C1
But, since all the charges are same,
1
1
1


(series)
C total C1 C 2
Example 7
8 μF
C1
24 V
4 μF
6 μF
(a) Find the equivalent capacitance of the capacitors above.
(b) Determine the total charge in the circuit.
14
2 μF
(c) Determine the charge on one plate of C1.
(d) Determine the electrical energy stored in C1.
Solution
(a) We can simply add the 6 F and the 2 F capacitors, since they are in parallel, 6 F +
2F = 8 F. But since the two 8 F capacitors are in series, we must add their inverses to
find their equivalent capacitance:
1
1
1
2



C88 8F 8F 8F
C88  4 F
Now the two 4 F capacitors are in parallel with each other, and can be added:
CT  4F  4F  8F
(b) qtot  CT   8F 24V   192 C
(c) C1 is in parallel with the battery and thus has 24 V across it.
qtot  C1  4F 24V   96 C
(d) U 
1
1
1
2
C1V 2 1  C1 2  4F 24 V   1152J
2
2
2
RC Circuits
A resistance-capacitance (RC) circuit is simply a circuit containing a battery, a resistor,
and a capacitor in series with one another. An RC circuit can store charge, and release it
at a later time. The capacitive time constant c is equal to the product of R and C (which
has units of time), and gives an indication of how long it takes for a capacitor to charge or
discharge. Typically, when connected to a battery, a capacitor is fully charged in a time
of approximately 5RC.
a
R
I
b
+
ε
C
-
When the switch is moved to position a, current begins it flow from the battery, and the

capacitor begins to fill up with charge. Initially, the current is
by Ohm’s law, but then
R
decreases as time goes until the capacitor is full of charge and will not allow any more
charge to flow out of the battery.
15
This leads us to a couple of rules we can follow when dealing with capacitors in an RC
circuit:
1. An empty capacitor does not resist the flow of current, and thus acts like a wire.
2. A capacitor which is full of charge will not allow current to flow, and thus acts like a
broken wire.
If we move the switch to position b, the battery is taken out of the circuit, and the
capacitor begins to drain its charge through the resistor, creating a current in the opposite
direction to the current flowing when the battery was connected.
a
R
I
b
+
ε
C
-
Eventually, the current will die out because of the heat energy lost through the resistor.
Example 8
a
3Ω
I
b
+
12V
6μF
-
A 12-volt battery is connected in series to a 3  resistor and an initially uncharged
capacitor. The switch is connected to position a in the circuit shown above.
(a) Determine the current in the circuit immediately after the battery is connected to the
resistor and capacitor,
(b) Determine the current in the circuit a long time later.
(c) Determine the maximum charge on the capacitor.
The switch is then moved to position b.
(d) Determine the total amount of energy dissipated through the resistor.
Solution
(a) Immediately after the circuit is connected, the capacitor is still empty and thus acts
like a wire.
16
The current, then, is I 
V 12V

 4A
R 3
(b) Then the current begins to decrease as the capacitor fills up with charge, and after a
long time, the capacitor is full of charge, and the current stops flowing completely. Thus,
I = 0 a long time later.
(c) When the capacitor is full of charge, the voltage across it is equal and opposite to the
battery emf.
q  C  6F 12V   72 C
(d) The total energy dissipated through the resistor is equal to the maximum electrical
energy UE stored in the capacitor:
1
1
1
2
U  CV 2  C 2  6F 12V   432 J .
2
2
2
REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
Questions 1 – 3:
(D) 2 A
(E) 1 A
Two resistors of 2  and 4 are placed
in series with a 12-V battery.
3. The voltage across the 4 resistor is
(A) 2 V
(B) 4 V
(C) 6 V
(D) 8 V
(E) 12 V
1. Which of the following statements is
true?
(A) The 2will get more current than
the 4 resistor since it has less
resistance.
(B) The 4 will get more current than
the 2 resistor since it has more
resistance.
(C) The 2  and 4  resistors will get
the same voltage.
(D) The 2 resistor will get more
voltage than the 4 resistor.
(E) The 4 resistor will get more
voltage than the 2 resistor.
2. The current in the 2 resistor is
(A) 6 A
(B) 4 A
(C) 3 A
17
Questions 4 – 6
A1
6. What is the correct reading on
ammeter 2?
(A) zero
(B) 1 A
(C) 2 A
(D) 3 A
(E) 6 A
A2
3Ω
V1
6Ω
6V
2Ω
2Ω
A3
4V
2Ω
V2
Two resistors of 3 and 6 are placed
in parallel with a 6-V battery. Three
ammeters and two voltmeters are placed
in the circuit as shown.
7. What is the total current flowing in
the circuit shown above?
(A) 4 A
(B) 3 A
(C) 2 A
(D) 4/3 A
(E) 2/3 A
4. Which of the following statements is
true of the voltmeters?
(A) Voltmeter 1 and voltmeter 2 will
read the same voltage.
(B) Voltmeter 1 will read 6 V.
(C) Voltmeter 2 will read 6 V.
(D) Voltmeter 2 will read the correct
voltage across the 3 resistor.
(E) Both voltmeters will read the correct
voltage across the 6 resistor.
2Ω
C
10V
8Ω
5. Which ammeter will read the highest
amount of current?
(A) Ammeter 1
(B) Ammeter 2
(C) Ammeter 3
(D) All three will read the same current.
(E) All three will read zero current.
8. A resistance-capacitance circuit is
connected as shown above.
After a long time, the current in the
2resistor is
(A) zero
(B) 10 A
(C) 8 A
(D) 2 A
(E) 1 A
18
2A
10. The aluminum cylinder above has a
length L and a cross-sectional area A.
Which of the following would double
the resistance of the aluminum?
(A) double both the length and the area
(B) double only the length
(C) double only the area
(D) reduce the length by half
(E) reduce the area by one quarter
6Ω
20V
9. The internal resistance of the battery
in the portion of the circuit above is
(A) 12 Ω
(B) 10 Ω
(C) 8 Ω
(D) 7 Ω
(E) 4 Ω
Free Response Question
Directions: Show all work in working the following question. The question is worth 15 points,
and the suggested time for answering the question is about 15 minutes. The parts within a
question may not have equal weight.
1. (15 points)
4Ω
ε=6V
2Ω
6Ω
3 μF
The switch in the circuit above is initially open and the capacitor is uncharged. The
switch is then closed.
(a) Determine the total current in the circuit immediately after the switch is closed.
(b) Determine the current in the circuit after the switch has been closed for a long time.
19
(c) On the axes below, sketch a graph of current I vs time t from the time the switch is
closed at t = 0 to the time when the capacitor is completely full of charge.
Be sure to label any important points on the axes.
I
t
0
(d) Determine the voltage across the capacitor after it is fully charged.
(e) After a long time, the 2 Ω resistor fails, breaking the circuit at the location of the 2 Ω
resistor, but before the parallel combination of the 6 Ω resistor and the capacitor.
Determine the magnitude and direction of the initial current in the 6 Ω resistor
immediately after the 2 Ω resistor fails.
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. E
Since both resistors get the same current, the 4 resistor needs more voltage than the 2
resistor, since the 4 has more resistance.
2. D
The total resistance in the series circuit is the sum of the resistors, or 6.
The current in the 2 resistor is the same as the total current in the circuit, which is I =
V/R = 12V/6 = 2 A.
3. D
Since the 4 resistor represents 2/3 of the total resistance, it will get 2/3 of the total
voltage, or 8 V.
4. B
The voltage across the 6 resistor is 6 V since it is in parallel with the battery. Voltmeter
2 is incorrectly placed in the circuit and will read zero volts.
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5. A
Ammeter 1 is in a position to measure the total current in the circuit before it splits up to
go through the 3 and 6 resistors.
6. B
Ammeter 2 measures only the current through the 6 resistor: I = 6V/6 = 1 A.
7. B
1
1
1
3



, which implies that Rt = 4/3 
Rt 2 4 4
V
4V
So, the total current in the circuit is I  
 3A .
R 4

3
8. E
After a long time, the capacitor is full of charge and does not allow any more current to
flow through it. Thus, the current from the battery will only flow through the resistors
and the current will be I = 10V/10 = 1 A.
The total resistance can be found by
9. E
By Ohm’s law, 2 A 
20V
, so the internal resistance R = 4 Ω.
6  R 
10. B
L
 2 L 
R
, so R 
 2R
A
A
Free Response Question Solution
(a) 3 points
Immediately after the switch is closed the capacitor is empty and acts like a wire (short)
around the 6 Ω resistor. Thus, the current in the circuit is
V
6V
Io 

1A
Rtot 6
(b) 2 points
After a long time, the capacitor is full, so no current flows through it, and the current
flows through the 6 Ω resistor.
V
6V
If 

 0.5 A
Rtot 12
(c) 4 points
I
1A
0.5A
21
t
0
(d) 3 points
When the capacitor is fully charged, it has the same voltage across it as the 6 Ω resistor,
since they are in parallel.
Vc  V6  I f R6  0.5 A6   3V
I
(e) 3 points
The capacitor acts as the source of voltage for
the current through the 6 Ω resistor.
3V
I
 0.5 A counterclockwise
6
22
6Ω
3 μF
3 μF
MAGNETIC FORCES AND MAGNETIC FIELDS
A magnetic field is the condition of the space around a magnet in which another magnet
will experience a force. Magnetic poles can be north or south, and like poles repel each
other and unlike poles attract. Fundamentally, magnetism is caused by moving charges,
such as a current in a wire. Thus, a moving charge or current-carrying wire produces a
magnetic field, and will experience a force if placed in an external magnetic field.
QUICK REFERENCE
Important Terms
electromagnet
a magnet with a field produced by an electric current
law of poles
like poles repel each other and unlike poles attract
magnetic domain
cluster of magnetically aligned atoms
magnetic field
the space around a magnet in which another magnet or moving charge
will experience a force
mass spectrometer
a device which uses forces acting on charged particles moving a magnetic field
and the resulting path of the particles to determine the relative masses of the
charged particles
right-hand rules
used to find the magnetic field around a current-carrying wire or the
force acting on a wire or charge in a magnetic field
solenoid
a long coil of wire in the shape of a helix; when current is passed through a
solenoid it produces a magnetic field similar to a bar magnet
23
Equations and Symbols
B
FB
q 0 v sin 
where
mv
qB
FB  ILB sin 
r
B
B = magnetic field
FB = magnetic force
q = charge
v = speed or velocity of a charge
θ = angle between the velocity of a
moving charge and a magnetic field,
or between the length of a currentcarrying wire and a magnetic field
r = radius of path of a charge moving in
a magnetic field, or radial distance
from a current-carrying wire
m = mass
I = current
L = length of wire in a magnetic field
μ0 = permeability constant
= 4π x 10-7 (T m) / A
0 I
2r
The Force That a Magnetic Field Exerts on a Moving Charge, The
Motion of a Charged Particle in a Magnetic Field, and The Mass
Spectrometer
Since a moving charge creates a magnetic field around itself, it will also feel a force
when it moves through a magnetic field. The direction of the force acting on such a
charge is given by the right-hand rule, with the thumb pointing in the direction of the
velocity of the charge. We use our right hand for moving positive charges, and our left
hand for moving negative charges.
F
F
N
S
v
B
I or v
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B
Right-hand Rule No. 1 for force on a moving charge: Place your fingers in the
direction of the magnetic field (north to south), your thumb in the direction of the velocity
of a moving charge (or current in a wire), and the magnetic force on the charge (or wire)
will come out of your palm.
The equation for finding the force on a charge moving through a magnetic field is
F = qvBsin 
where q is the charge in Coulombs, v is the velocity in m/s, B is the magnetic field in
Teslas, and  is the angle between the velocity and the magnetic field. If the angle is 90,
the equation becomes F = qvB.
Example 1
A proton enters a magnetic field B which is directed into the page. The proton has a
charge +q and a velocity v which is directed to the right, and enters the magnetic field
perpendicularly.
B
q = +1.6 x 10-19 C
v = 4.0 x 106 m/s
q
B = 0.5 T
v
Determine
(a) the magnitude and direction of the initial force acting on the proton
(b) the subsequent path of the proton in the magnetic field
(c) the radius of the path of the proton
(d) the magnitude and direction of an electric field that would cause the proton to
continue moving in a straight line.
Solution
(a) As the proton enters the magnetic field, it will initially experience a force which is
directed upward, as we see from using the right-hand rule.
B
F
q

v


F  qvB  1.6 x10 19 C 4.0 x10 6 m / s 0.5T   3.2 x10 13 N
25
(b) The path of the proton will curve upward in a circular path, with the magnetic force
becoming a centripetal force, changing the direction of the velocity to form the circular
path.
(c) The radius of this circle can be found by setting the magnetic force equal to the
centripetal force:
magnetic force = centripetal force
mv 2
qvB 
r
mv 1.7 x10 27 kg 4.0 x10 6 m / s
r

 8.5 x10 2 m
19
qB
1.6 x10 C 0.5T 





(d) If the charge is to follow a straight-line path through the magnetic field, we must
orient the electric field to apply a force on the moving charge that is equal and opposite to
the magnetic force. In this case, the electric force on the charge would need to be directed
downward to counter the upward magnetic force. The electric field between the plates
would be directed downward, as shown below:
B
q
v
The net force acting on the moving charge is the sum of the electric and magnetic forces,
and is called the Lorentz force:
Fnet  FE  FB
Fnet  qE  qvB
The net force in this case is zero, so the magnitude of the electric and magnetic forces are equal to each other :
qE  qvB


E  vB  4.0 x10 6 m / s 0.5T   2.0 x10 6
N
C
This expression relates the speed of the charge and the electric and magnetic fields for a
charge moving undeflected through the fields.
26
The Force on a Current in a Magnetic Field
Since a current-carrying wire creates a magnetic field around itself according to the first
right-hand rule, every current-carrying wire is a magnet. Thus, if we place a currentcarrying wire in an external magnetic field, it will experience a force. Once again, the
direction of the force acting on the wire is given by the right-hand rule:
F
N
S
B
B
I or v
I
Again, you would use your left hand to find the direction of the magnetic force if you
were given electron flow instead of conventional current.
The equation for finding the force on a current-carrying wire in a magnetic field is
F = ILBsin 
where I is the current in the wire, L is the length of wire which is in the magnetic field, B
is the magnetic field, and  is the angle between the length of wire and the magnetic
field. If the angle is 90 , the equation becomes simply F = ILB.
Example 2
A wire carrying a 20 A current and having a length L = 0.10 m is placed between the
poles of a magnet at an angle of 45, as shown. The magnetic field is uniform and has a
value of 0.8 T.
Top View
45
N
S
B
I
Determine the magnitude and direction of the magnetic force acting on the wire.
(sin 45 = cos 45 = 0.7)
27
Solution
The magnitude of the force on the wire is found by
F  ILB sin   20 A0.10 m0.8T sin 45  1.13 N
The direction of the force can be found by the right-hand rule. Place your fingers in the
direction of the magnetic field, and your thumb in the direction of the length (and current)
which is perpendicular to the magnetic field, and we see that the force is out of the page.
Note that the length must have a component which is perpendicular to the magnetic field,
or there will be no magnetic force on the wire. In other words, if the wire is placed
parallel to the magnetic field, sin 0 = 0, and the force will also be zero.
Remember, use your right hand for current or moving positive charges, and your left
hand for electron flow or moving negative charges.
Example 3
A wire is bent into a square loop and placed completely in a magnetic field B = 1.2 T.
Each side of the loop has a length of 0.1m and the current passing through the loop is 2.0
A. The loop and magnetic field is in the plane of the page.
(a) Find the magnitude of the initial force on
each side of the wire.
(b) Determine the initial net torque acting on the
loop.
B
a
b
I
c
d
Solution
By the right-hand rule, side ab will experience a force downward into the page and side
cd will experience a force upward out of the page. The current in sides bd and ac are
parallel to the magnetic field, so there is no magnetic force acting on them.
Fab  ILB  2.0 A0.1m1.2T   0.24 N into the page.
Fcd  ILB  2.0 A0.1m1.2T   0.24 N out of the page.
The result of the opposite forces on ab and cd is a torque on the loop, causing it to rotate
in the magnetic field. This is the basic principle behind ammeters, voltmeters, and the
electric motor. In this case, two equal and opposite forces cause the torque on the loop:
  2rF  20.05 m0.24 N   0.024 Nm
28
Magnetic Fields Produced by Currents, and Ampere’s Law
A current-carrying wire creates a magnetic field around itself. Fundamentally, magnetic
fields are produced by moving charges. This is why all atoms are tiny magnets, since the
electrons around the nucleus of the atom are moving charges and are therefore magnetic.
The magnetic field due to a current-carrying wire circulates around the wire in a direction
can be found by another right-hand rule.
Right-hand Rule No. 2 for the magnetic field around a current-carrying wire:
Place your thumb in the direction of the current I, and your fingers will curl around in
the direction of the magnetic field produced by that current.
current I
I
Magnetic Field B
In determining the direction of a magnetic field due to the flow of electrons in a wire, we
would use the left hand instead of the right hand.
If the distance r from the wire is small compared to the length of the wire, we can find the
magnitude of the magnetic field B by the equation
B
o I
2r
r
I
Tm
. The value 4
A
in this constant is often used for reasons of geometry. The magnetic field around a
1
current-carrying wire is proportional to , while electric field around a point charge is
r
1
proportional to 2 .
r
where o is called the permeability constant and is equal to 4 x 10-7
29
REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
3. The magnitude of the force on the
wire is
(A) 0.06 N
(B) 2.0 N
(C) 6.7 N
(D) 0.15 N
(E) 0.015 N
1. A wire carries a current, creating a
magnetic field around itself as shown.
The current in the wire is
(A) directed to the right.
(B) directed to the left.
(C) equal to the magnetic field.
(D) in the same direction as the magnetic
field.
(E) zero.
Questions 4 – 6:
An electron enters a magnetic field as
shown.
e
B
Questions 2 – 3:
A wire carrying a current of 2 A is
placed in a magnetic field
of 0.1 T as shown. The length of wire in
the magnetic field
is 0.3 m.
θ
4. The electron will experience a force
which is initially
(A) into the page.
(B) out of the page.
(C) toward the top of the page.
(D) toward the bottom of the page.
(E) to the left.
B
N
v
5.The magnitude of the force acting on
the electron is
(A) evB
(B) evB cos
(C) evBsin 
(D) vB
(E) vBsin 
S
I
6. The resulting path of the electron is a
(A) parabola
(B) straight line
(C) spiral or helix
(D) hyperbola
(E) circle
2. The force on the wire is directed
(A) into the page.
(B) out of the page.
(C) toward the top of the page.
(D) toward the bottom of the page.
(E) to the left.
30
Free Response Question
Directions: Show all work in working the following question. The question is worth 10 points,
and the suggested time for answering the question is about 10 minutes. The parts within a
question may not have equal weight.
1. (10 points)
P
y
r
2r
I
x
+z
(out of page)
4I
Two wires cross each other at right angles. The vertical wire is carrying a current I and the
horizontal wire is carrying a current 4I. Point P is a perpendicular distance r from the vertical
wire, and a distance 2r from the horizontal wire.
(a) With reference to the coordinate system shown at the right, determine the magnitude and
direction of the magnetic field at point P.
P
r
v
e
I
2r
4I
An electron is moving parallel to the horizontal wire with a speed v in the +x direction.
Determine each of the following as the charge passes point P:
(b) the magnitude and direction of the net force acting on the electron
(c) the magnitude and direction of the electric field necessary to keep the electron moving in a
straight horizontal path.
31
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. A
By right-hand rule no. 2, if the fingers curl in the direction of the magnetic field around the wire
(up and over the wire toward you), the current must be to the right.
2. B
According to right-hand rule no. 1, the fingers point to the right in the direction of the magnetic
field, the thumb points toward the bottom of the page in the direction of the current, and the force
comes out of the palm and out of the page.
3. A
F = ILB = (2 A)(0.1 T)(0.3 m) = 0.06 N
4. B
We use the left-hand-rule, since the electron is a negative charge, placing our fingers toward the
bottom of the page, and the thumb to the right, in the direction of the component of the velocity
which crosses the magnetic field lines. The force, then, comes out of the palm and out of the
page.
5. C
The magnitude of the force is equal to the product of the charge, speed, and magnetic field, and
the sine of the angle between the velocity and the magnetic field, which is θ.
6. C
The electron will orbit a magnetic field line, but also continue to move down toward the bottom
of the page, therefore spiraling downward.
Free Response Question Solution
(a) 4 points
The net magnetic field at point P is due to the magnetic fields produced by both wires:
 I
Br  o into the page (-z)
2r
 4 I 
out of the page (+z)
B2 r  o
2 2r 
 4 I   o I  o I
Bnet  B2 r  Br  o


out of the page (+z)
2 2r  2r 2r
(b) 2 points
Let the charge on one electron be e. Then
 I 
F  evB  ev o 
 2r 
The direction of the force is up to the top of the page (+y) by the left-hand rule for a moving
negative charge.
32
(c) 4 points
In order to keep the electron moving in a straight path to the right, we would need to apply a
downward (-y) electric force to the electron which is equal and opposite to the upward (+y)
magnetic force.
Since electric field lines are drawn in the direction a
positive charge would experience a force, they would
be drawn in the opposite direction an electron would
experience a force. Thus, the electric field would need
to be applied in the +y direction to keep the electron
moving in a straight horizontal line.
FB
E
FE
The magnitude of the electric field can be found by setting the magnetic force equal to the
electric force:
FB  FE
evB  eE
 I 
E  vB  v o 
 2r 
33