Unit 17
Magnetic Flux and Faraday’s Law of Induction
17.1
Ampere’s law
17.2
Forces between current wires
17.3
Magnetism
17.4
Induced EMF
17.5
Magnetic flux
17.6
Faraday’s law of induction
17.7
Lenz’s law
17.8
Motional EMF
17.1 Ampere’s law
Electric currents can create magnetic fields. The direction of the
magnetic field is given by the magnetic field right-hand rule. The
magnetic field right-hand rule states the following:
To find the direction of the magnetic field due to a current-carrying
wire, point the thumb of your right hand along the wire in the
direction of the current I. Your fingers are now curling around the
wire in the direction of the magnetic field.
Experiments show that the field produced by a current-carrying wire
doubles if the current I is doubled. In addition, the field decreases by a
factor of 2 if the distance from the wire, r, is doubled. Hence, we
conclude that the magnetic field B must be proportional to I / r ; that is
B∝
I
.
r
d d
Ampere’s law states that the sum of B ⋅ dl along the closed path is
proportional to the current enclosed by the path. Mathematically, we
have
d d
B
∫ ⋅ dl = µ 0 I enclosed .
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The proportional constant, µ0 is the permeability of free space. Its value is
m 0 = 4π × 10 −7 T ⋅ m / A .
For example, the magnetic field at a distance r due to a long wire of current can be obtained
by the Ampere’s law.
d
d
∫ B ⋅ dl
which gives B =
= B (2πr ) = µ 0 I enclosed ,
µ0 I
(the same as that obtained by the Biot and Savart’s law).
2π r
Remark:
Recall that in the long solenoid, which has n turns per unit
length of solenoid and carries a current I, the magnetic
field B at a point O on the axis of solenoid is found to be
B = µ 0 nI .
In fact, this expression can be obtained simply by using
the Ampere’s law. Consider the amperian loop as shown
in figure. The length, side 1 of the amperian loop is L,
which has N turns of coils. The magnetic field is nearly
uniform and tightly packed inside the loops. In the ideal
case of a very long, tightly packed solenoid, the
magnetic field outside is practically zero – especially
when compared with the intense field inside the solenoid.
We can use this idealization, in combination with
Ampere’s law, to calculate the magnitude of the field inside the solenoid.
d d
B
∫ ⋅ dl = ∑ B// ∆L + ∑ B// ∆L + ∑ B// ∆L + ∑ B// ∆L = BL + 0 + 0 + 0 = µ 0 I enclosed
side 1
side 2
side 3
side 4
The answer is simply BL = µ 0 IN = µ 0 I (nL) , which gives B = µ 0 nI .
Example
If you want to increase the strength of the magnetic field inside a solenoid is it better to (a)
double the number of loops, keeping the length the same, or (b) double the length, keeping
the number of loops the same?
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Answer
Since B = µ0 nI , we know that the number of coil per unit length of solenoid governs the
magnitude of magnetic field. Doubling the number of loops and keeping the length the same,
results in doubling the variable n. So, the answer is (a).
Example
Two wires separated by a distance of 22 cm carry currents in the same direction. The current
in one wire is 1.5 A, and the current in the other wire is 4.5 A. Find the magnitude of the
magnetic field halfway between the wires.
Answer
The magnetic field due to I1:
=
B
1
mm
0 I1
0 (1.5 A)
=
= 2.7 × 10−6 T , into page.
−2
2π (r / 2) 2π (11.0 × 10 m)
The magnetic field due to I2:
B=
2
mm
0 I2
0 (4.5 A)
=
= 8.2 × 10−6 T , out of page.
−2
2π (r / 2) 2π (11.0 × 10 m)
The net magnitude of magnetic field: B = B2 − B1 = 5.5 × 10−6 , out of page.
17.2 Forces between current wires
We know that current-carrying wire in a magnetic field experiences
a force. We also know that a current-carrying wire produces a
magnetic field. It follows, then, that one current-carrying wire will
exert a force on another.
The force experienced by wire 2 in the figure, has a magnitude
 µ0 I1  µ0 I1 I 2
=
F I=
I2 L  =
L.
2 LB

2π d
 2π d 
Similarly, we notice that an equal magnitude but opposite direction force
is experienced by wire 1. To conclude,
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Wires with parallel currents attract one another.
Wires with opposite currents repel one another.
Remark:
When the magnitude of the force per unit length between two infinitely long, parallel wires
that carry identical currents and are separated by 1 m is 2 × 10−7 N/m, the current in each wire
is defined to be 1 A.
17.3 Magnetism
There are three types of magnetism.
(a)
Ferromagnetism – A ferromagnetic material produces a magnetic field even in the
absence of an external magnetic field. Permanent magnets are constructed of
ferromagnetic materials, e.g. bar magnets.
(b)
Paramagnetism – A paramagnetic material has no magnetic field unless an
external magnetic field is applied to it. In this case it develops a magnetization in
the direction of the external field.
(c)
Diamagnetism – Diamagnetism is the effect of the production by a material of a
magnetic field in the opposite direction to an external magnetic field that is
applied to it. All materials show at least a small diamagnetic effect.
17.4 Induced EMF
A simple experiment was demonstrated by Faraday to observe the induced emf. In the
experiment, two electrical circuit are involve. The first, called the primary circuit, consist of a
battery, a switch, a resistor to control the current, and a coil of several turns around an iron
bar. When the switch is closed on the primary circuit a current flows through the coil,
producing a magnetic field that is particularly intense within the iron bar. The second circuit
also has a coil wrapped around the iron bar, and this coil is connected to an ammeter to detect
any current in the circuit. Note, however, that there is no direct physical contact between the
two circuits.
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When the switch is closed on the primary circuit the magnetic field in the iron bar rises from
zero to some finite amount, and the ammeter in the secondary coil deflects to one side briefly,
and then returns to zero. As long as the current in the primary circuit is maintained at a
constant value the ammeter in the secondary circuit gives zero reading. If the switch on the
primary circuit is now opened, so that the magnetic field decreases again to zero, the ammeter
in the secondary circuit deflects briefly in the opposite direction, and then returns to zero. The
current in the second circuit is referred to as the induced current, because the two circuit has
no direct contact. As we know that there should be an induced emf to produce such an
induced current. However, the changing magnetic field is caused by a changing current in the
primary circuit. The following demonstration shows the same idea.
17.5 Magnetic flux
The magnetic flux Φ is defined as the dot product of the magnetic field and the area of the
 
current loop, e.g. Φ = B ⋅ A , in scalar representation, we have Φ = BA cosθ , where θ is the
angle between the magnetic field B and the unit vector of the area.
The SI unit of Φ is weber, where 1 weber = 1 Wb = 1 T⋅m2.
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Example
The three loops of wire shown in the figure are all in a region of
space with a uniform, constant magnetic field. Loop 1 swings back
and forth as the bob on a pendulum; loop 2 rotates about a vertical
axis; and loop 3 oscillates vertically on the end of a spring. Which
loop or loops have a magnetic flux that changes with time?
Answer
Loop1 moves back and forth, and loop 3 moves up and down, but since the magnetic field is
uniform, the flux does not depend on the loop’s position. Loop 2, on the other hand, changes
its orientation relative to the field as it rotates; hence, its flux does change with time. The
answer is loop 2.
17.6 Faraday’s law of induction
Faraday found that the second coil in the experiment described above experiences an induced
emf which is given by the following relation:
E = −N
dΦ
dt
This is known as faraday’s law of induction, the minus sign in the right of the expression
indicates that the induced emf opposes the change in magnetic flux. The variable N is the
number of loops in a coil. The above relation can be written simply as E = − N
∆Φ
, if the
∆t
change in magnetic flux is uniform with time.
Example
A bar magnet is moved rapidly toward a 40-turn, circular coil of wire. As the magnet moves,
the average value of B cosθ over the area of the coil increases from 0.0125 T to 0.450 T in
0.250 s. If the radius of the coil is 3.05 cm, and the resistance of its wire is 3.55 Ω, find the
magnitude of (a) the induced emf and (b) the induced current.
Answer
The cross sectional area of the coil:=
A π=
r 2 π (0.0305 m) 2 .
The initial magnetic flux is given by
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2
3.65 × 10−5 T ⋅ m 2 .
Φ=
Bi =
A (0.0125 T )π (0.0305 m)=
i
The final magnetic flux is given by
Φ f= B f A= (0.450 T )π (0.0305 m) 2= 1.32 × 10−3 T ⋅ m 2 .
Applying the Faraday’s law, we have
1.32 × 10−3 T ⋅ m 2 − 3.65 × 10−5 T ⋅ m 2
∆Φ
=
| E | N= (40)
= 0.205V .
∆t
0.250 s
Use the Ohm’s law to calculate the induced current: =
I
V 0.205V
=
= 0.0577 A .
R 3.55Ω
Remark:
If the magnet is now pulled back to its original position in the same amount of time, the
induced emf and current will have the same magnitudes; their directions will be reversed.
17.7 Lenz’s law
Lenz’s law states that an induced current always flows in a direction that opposes the change
that caused it. As an illustration, consider the magnetic field which decreases with time. The
induced current flows through the ring so as to oppose this change, producing a magnetic
field within the ring in the same direction as the decreasing magnetic field.
Example
If the north pole of a magnet is moved toward a conducting loop, the induced current
produces a north pole pointing toward the magnet’s north pole. This creates a repulsive force
opposing the change that caused the current. On the other hand, if the north pole of a magnet
is pulled away from a conducting loop, the induced current produces a south magnetic pole
near the magnet’s north pole. The result is an attractive force opposing the motion of the
magnet.
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Answer
According to Lenz’s law, the induced current in the ring produces a magnetic field that exerts
a repulsive force on the magnet – to oppose its motion. In contrast, the ring on the right has a
break, so it cannot have a circulating current. As a result, it exerts no force on its magnet.
Therefore, the magnet in the left falls down with an acceleration smaller than the
gravitational acceleration. But, the right magnet falls with the gravitational acceleration.
Consider a system in which a metal ring is falling out of a region with a magnetic field and
into a field-free region, as shown in the figure. According to Lenz’s law, the induced current
in the ring is counterclockwise. The reasons are as follows. The
induced current must be in a direction that opposes the change in
the system. In this case, the change is that fewer magnetic field
lines are piercing the area of the loop and pointing out of the page.
The induced current can oppose this change by generating more
field lines out of the page within the loop. This can be
accomplished
by
an
induced
current
circulating
in
a
counterclockwise direction.
Note also that an upward force is experienced at the top of the ring. No force is experienced
at the bottom of ring.
The retarding effect on a ring leaving a magnetic field, allows us to
understand the behavior of eddy currents. This is also known as the
magnetic braking. There are several points worth notice.
•
There is no direct physical contact, thus eliminating frictional wear.
•
The magnetic braking force is stronger if the speed of the metal is
greater.
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The same idea applies to a metal sheet which is leaving the magnetic field. Eddy current
flows within the metal sheet and its direction is such that the change is opposed.
17.8 Motional EMF
A conducting rod of length l is moving with constant speed v, as shown in figure. The change
in magnetic flux ∆Φ in a time interval ∆t is given by
∆Φ = B∆A = B (vl∆t ) .
Applying Faraday’s law, we have
 ∆Φ 
 Bvl∆t 
E = N
 = (1)
 = Bvl
 ∆t 
 ∆t 
The current in the conducting rod is I =
E Bvl
. The
=
R
R
magnetic force experienced by the conducting rod is
2
2
 Bvl  B vl
=
Fmagnetic = BIl = B
l

R
 R 
.
So,
an
equal
magnitude but opposite direction force, that is Fexternal, must be applied to maintain a
movement of it with a constant speed.
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