MAGNETIC INDUCTION AND FARADAY-LENZ LAW
In a previous chapter, we saw that if we put a closed conducting loop in a magnetic field
and then send a current through the loop, the field exerts a torque that turns the loop:
current loop  magnetic field  torque.
Suppose that, instead, with the current off, we turn the loop by hand. Will a current now
appear in the loop? The answer is yes.
torque  magnetic field  current.
The law on which this phenomenon is based is
called Faraday's law of induction.
The figure on the right shows a conducting loop
connected to an ammeter. There is no battery and
hence no current. However, if we move a bar
magnet toward the loop, a current suddenly appears
in the loop. If we move the magnet away, a current
again appears, but in the opposite direction. If we
experimented for a while, we'd discover the
following:
(1) A current appears only if there is relative motion between the loop and the magnet.
(2) Faster motion produces a greater current.
(3) If moving the north pole toward the loop causes, say, a CW (clockwise) current, then
moving the north pole away causes a CCW current.
Moving the south pole toward or away also causes
currents, but CCW for "toward" and "CW" for "away".
The current produced in the loop is called an induced
current. The associated emf is called an induced emf.
The phenomenon is called induction.
Faraday's Law: Faraday realized that an emf and a
current can be induced in a loop by changing the amount of magnetic field passing
through the loop. The amount of field can be visualized in terms of the field lines
passing through the loop. Thus: An emf is induced in the loop when the number of field
lines through the loop is changing.
To put Faraday's law on a quantitative footing,
we introduce a quantity called magnetic flux
(  ) as a measure of the amount of magnetic
field:
Defn of Magnetic Flux:
 
   B·dA
Suppose an arbitrarily shaped surface is immersed in a magnetic field. Consider an
element of area dA . The flux through this element is d   BdA cos  .
Consider the special case of a plane of

area A in a uniform field B that makes

a constant angle  with A . The
magnetic flux through the plane in this
case is:
  BA cos 
 
If B  A , as in Fig (b), then   0 , and

the flux is  max  BA . If B is parallel
to the plane, as in Fig (a), then   90 ,
and   0 . The unit of flux is the
weber (Wb): 1 Wb= 1 T.m2.
Example A circular area of radius 6.50 cm lies in the xy -plane. What is the magnetic
flux through this circle due to a uniform magnetic field B  0.230 T (a) in the
 z direction? (b) at angle of 53.1 from the  z direction? (c) in the  y direction?


Solution: (a) B and A are parallel, so the flux is
 
  B·A  BA  (0.23) (0.065) 2  3.05 10 3 Wb  3.05 mWb
 
(b)   B·A  BA cos  (0.23) (0.065) 2 cos53.1  1.83 mWb


(c) B and A are perpendicular, so the flux is
 
  B·A  BA cos 90  0.
Example Flux through a rectangular loop: A loop of length b and width a is located
near a long wire carrying a current I . The distance between the wire and the closer side
of the loop is c . Find the flux through the loop.
Solution: The field of a long wire is
B  o / 2 r . This field varies over the
loop, so we cannot simply use   BA to
find the flux. We first find the flux
d  through an infinitesimal strip of length
b and width dr over which the field is
uniform. We integrate over r to find the
total flux. The strip is rectangular, so its
area is dA  bdr and the flux through it is
d   BdA  Bbdr .
ca
o I
o Ib c  a dr o Ib
o Ib  c  a 
ca
ln
ln 
   BdA  


bdr 
r




c




2
2
2
2
r
r
 c 
c
c
Faraday's Law: With the notion of flux, we can state Faraday's law in a more
quantitative and useful way: The magnitude of the emf  induced in a loop is equal to the
rate at which the flux  through that loop changes with time. It was found by Lenz that
the induced emf tends to oppose the change in flux, so we write Faraday's law as
d
(Faradays law)
 
dt
with the minus sign indicating that opposition. The minus sign is best understood using
Lenz's law (to be discussed next), so we often neglect it, seeking only the magnitude of
 . If we change the flux through a coil of N turns, then an induced emf appears in every
turn and so the total emf is
d
 N
(Faradays law for a coil of N turns)
dt
Since   BA cos  , the three ways in which we can change the flux are: (i) change B ;
(ii) change the area A ; (iii) change the angle  between the field and the area of the coil.
Exercise: The graph gives the magnitude
B (t ) of a uniform (but not constant) field that
exists throughout a conducting loop,
perpendicular to the plane of the loop. Rank
the regions according to the magnitude of the
emf induced in the loop, greatest first.
Solution: In regions a and c , there is no change in flux, hence there no emf is induced.
In regions d and e , the flux is changing at the same rate, so the emf will be the same in
these two regions. In region b , the flux is changing at a greater rate than in regions
d and e , so the greatest emf (magnitude) will be in region b .
Ans: b  d  e  a  c  0 .
Lenz's Law: Soon after Faraday proposed his law induction, Lenz proposed a rule-known as Lenz's law--for determining the direction of an induced current in a loop:
An induced current has a direction such that the magnetic field due to the induced current opposes the change in flux that induces the current.
In plain language, the law says the system creates a flux that tries to neutralize the change
in external flux. To create this flux, the system must have a current induced in it.
As an example, consider the experiment
(shown in the figure) where the north pole
of a magnet is moved toward a loop. This
motion induces a current in the loop. To
oppose the increase in flux, the loop
generates a field (and a magnetic moment

 ) directed rightward. (The field lines
shown in the figure are those of the

induced field Bi . The external field is
shown on the right.) The induced current
i runs CCW when viewed from the right.
The figure on the right depicts the field
lines of the moving bar magnet.
Please study the following set of situations
closely. A conducting loop is immersed in

an external field B with direction as indicated. Also shown are the directions of the


induced field Bi and the induced current i that produced it ( Bi ) in each case.
Note the opposition to flux change (stressed by Lenz's law) in each case. In each case,

verify (using RHR) that the direction of i is such that it produces the indicated Bi . For
example, in Fig (b) the flux of the external (leftward) field is decreasing, inducing a

current i that produces a field Bi leftward.
Exercise:
What is the direction of the induced current in the
loop due to the current shown in each of the
accompanying figures?
Electric guitars: An acoustic guitar depends for its sound on
the resonance produced in the hollow body of the instrument by
the oscillating strings. An electric guitar is a solid instrument,
so there is no body resonance. Instead, the oscillations of the
metal strings are sensed by electric "pickups" that send signals to
an amplifier.
The upper figure shows a side view of an electric guitar
pickup. The wire connecting the guitar to the amplifier is
coiled around a small magnet. The field of the magnet
produces a north and south pole in the section of the metal
string just above the magnet. When the string is plucked and
thus made to oscillate, its motion relative to the coil changes
the flux of its field through the coil, inducing a current in the
coil. The induced current oscillates at the same rate as the
string, thus relaying the frequency to the amplifier.
On a Fender Stratocaster (the Strat), there are three groups of
pickups, placed at the near end of the strings. The groups are
set up such that one group detects the high frequency range,
one the mid range, and the last the low range. By throwing a
toggle switch on the guitar, the musician can select which
group will send signals to the amplifier.
To gain further control over his music, Hendrix sometimes
rewrapped the wire in the pickup coils to change the number
of turns. In this way, he altered the induced emf and the
sensitivity to the oscillations of the strings.
Exercise: The figures show three situations in which identical circular loops are in

uniform B fields (  or  ) that are either
increasing (Inc) or decreasing (Dec) in
magnitude at identical rates. The dashed line
coincides with the diameter. In which case is
the induced current greatest (in magnitude)?
Is that current CW or CCW?
Three ways to change flux: Earlier we pointed that, since   BA cos  , the
three ways in which we can change the flux are: (i) change the field B ; (ii) change the
area A ; (iii) change the angle  between the field and the area of the coil.
We take these up in turn.
Example: Changing Magnetic Field
A circular loop of radius 10 cm has resistance
2.0  . The plane of the loop is perpendicular

to a uniform magnetic field B (  )that is
increasing at 0.10 T/s. Find the emf and
current induced in the loop.
Solution: The flux is given by   BA . The
magnitude of the emf induced is
d
dB
  r2
  (0.1) 2  0.1  3.14 mV.
dt
dt
The current magnitude is I  |  | / R  3.14 / 2.0  1.6 mA. The external field is into the
page and increasing. Therefore, by Lenz law, the induced field is directed out of the
page. To produce this field, the induced current must run CCW.
| | 
Example: Changing Magnetic Field
The figure shows a conducting loop consisting
of a semicircle o radius r  0.20 m and three
straight sections. The semicircle lies in a

uniform field B (  ); the field magnitude is
time-dependent: B  4.0t 2  2.0t  3.0 , where
SI units are implied. An ideal battery with emf
bat  2.0 V is connected to the loop. The
resistance of the loop 2.0  . What are the
magnitude and direction of the current induced in the loop at t  10 s?
Solution: We know the magnitude of the induced emf is ind  d  / dt . Let us find this
and evaluate it at t  10 s. The total emf in the circuit is the sum of the induced and
battery emfs.
d d
 r 2 dB  r 2 d
 r2
  ind 
 ( BA) 

(4t 2  2t  3) 
(8t  2)
2 dt
2 dt
2
dt
dt
At t  10 s ,
 (0.2) 2
 ind 
(8  10  2)  5.15 V.
2
The flux through the loop is out of the page and increasing. Therefore, the induced field
Bi must oppose this increase, and thus must be into page (  ). Using RHR, the induced
current must flow CW around the loop. ind must then also be CW. The two emfs, ind
and bat tend to drive currents in opposite directions. Because ind  bat , the net emf
net is also CW, and thus so is the current. The current magnitude at t  10 s is

  bat 5.15  2.0

 1.6 A.
I  net  ind
2.0
R
R
Example: Changing Area
Two parallel, horizontal, conducting rails are
a distance  apart. They are connected at one
end by a resistance R . A conducting bar
completes the loop, joining the two rails
electrically but free to slide along them.
Assume that the rails and the bar are of
negligible resistance. The whole circuit is
perpendicular to a uniform magnetic field

B (  ), as shown. (a) Let x be the
instantaneous position of the bar relative to
the fixed left end ( x  0 ). What is the flux
enclosed by the rectangle formed by the bar,
the rails, and the resistance R ? (b) An
external force ( Fext ) moves the bar to the right
at a constant speed v . What are the induced emf and current? (c) At what rate is
thermal energy being dissipated in the resistor R ? (d) What is the rate at which Fext
doing work on the bar? (e) What magnetic force ( FB ) is acting on the bar? What
relation holds between Fext and FB ?
 
Solution: (a)   B·A  BA  Bx.
(b) Let us get the magnitudes of the induced emf and current from Faraday's law. We'll
find the directions from Lenz's law. The magnitude of the induced emf is
d d
dx

 ( Bx)  B  Bv
dt
dt
dt
This is called a motional emf. From Ohm's law, the induced current is
 Bv
I 
R
R
Since the enclosed area is increasing, the flux is increasing; hence, by Lenz law, the
induced field must be out of the page (  ). The current must flow CCW to produce this
field.
(c) The rate of energy dissipation in R equals the power:
2
Bv
 Bv 
.
PR  I R  
R

R
 R 
(d) The rate at which Fext is doing work is
 
Pext  Fext ·v  Fext v,
Fext is not known. We instead appeal to energy conservation, by which Pext  PR .
2 2 2
2
 Pext  PR 
B 22v2
R
 Fext 
Pext B 2  2 v

.
v
R
 
(e) The magnetic force is given by I   B . The (induced) CCW current flows "up" the
bar and the external field is into the page (  ). Hence, FB  I B and is directed left.
Note that
B 2 2v
Fext  FB 
.
R
This makes sense: since the bar is moving at constant velocity, the net force on it is zero,
and Fext  FB .
Data: B  0.6 T, v  8 m/s,   15 cm, and R  25  . Evaluate the quantities in parts (a)
to (e) above.
Answers:   Bv  0.72 V; I   / R  28.8 mA;
PR  Pext  I 2 R  20.7 mW; Fext  FB  I B  2.59 mN.
Origin of the Motional emf: We can understand the
origin of the motional emf in terms of the magnetic force on
moving charges. The figure shows an electron in a conducting

rod of length  moving with a constant velocity v through a

uniform magnetic field B (  ). Because the electron is

moving with the same velocity v , there is a magnetic force
FB  evB directed downward as shown. Because of this force, free electrons move to the
lower end, and this end acquires a net negative charge. The lack of electrons at the upper
end gives it a net positive charge. This accumulation doesn't proceed forever: there is
now an electric field pointing down, and hence an upward force FE  eE on each
electron, and this force eventually balances FB . That is, eE  evB when equilibrium is
established. Cancelling e , we have E  vB . The potential difference across the rod is
V  E   vB . (The upper end is at a higher potential
than the lower end.) We identify this as the motional
emf   Bv that we obtained from Faraday's law. No
current flows in the moving rod, since no closed path
(circuit) is available. If the rod is moved on conducting
rails as shown, then a current I  Bv / R flows. The
direction of the current is CCW. (The electron flow is
CW.) Thus, we have shown that Faraday and Lenz laws are consistent with our previous
results.
Example: Non-uniform field
The figure shows a rod of length L caused to move at

constant velocity v along horizontal conducting rails.
The field in which the rod moves is provided by a
current i in a long wire parallel to the rails. Assume
that v  5.00 m/s, a  10.0 mm, L  10.0 cm, and
i  100 A. (a) Calculate the emf induced in the rod.
(b) The resistance of the rod is 0.400  and the rest of
the circuit is of negligible resistance. What is the
current induced in the loop? (c) At what rate is thermal
energy being generated in the rod? (d) What force must be applied to the rod by an
external agent to maintain the motion? (e) At what rate does this agent do work on the
rod?
Solution:
Example: Changing 
This is the principle behind an alternating current (ac) generator, a device that
converts mechanical energy to electrical energy. In its simplest form, a generator
consists of a loop of wire rotated by an external agent in a magnetic field.
In the figure, a rectangular loop is made to rotate with constant angular speed  about the

axis shown. The magnetic field B is uniform and constant. Find the induced emf.
Solution: The flux is given by
 
  B·A  BA cos   BA cos( o  t )  BA cos t taking  o  0.
The induced emf is
d
d
 
  BA cos  t   BA sin  t
dt
dt
The induced emf varies sinusoidally with time as shown in Fig (b). It does not depend on
the shape of the loop, but only on its area. Because  is directly proportional to  and
B , some tachometers use the emf in a rotating coil to measure rotational speed. Other
devices use an emf of this kind to measure magnetic field.
To produce a large emf, we use a coil of N turns, rather than a single loop. In this case,
d
d
 N
  N BA cos t  N  BA sin t.
dt
dt
If the resistance R of the coil is given, then we can find the current easily from
 NBA sin t
I 
.
R
R
The instantaneous power can be found from P  I 2 R or  2 / R . To find maximum
values, set t   / 2 .
Motors and generators are similar devices, just run in opposite ways. A motor converts
electrical energy to mechanical energy; a generator does the opposite. Often the same
device serves both purposes. In a hybrid car, for example, an electric motor takes energy
from a battery to provide propulsion. When the car brakes, the wheels turn the motor,
which then acts as a generator and puts the car's energy back into the battery instead of
dissipating it. Such regenerative braking is one of the several means by which a hybrid
car achieves greater efficiency.
Example: A rectangular coil of N turns
and of length a and width b is rotated at
frequency f in a uniform magnetic field

B (  ), as shown. The coil is connected to
co-rotating cylinders, against which metal
brushes slide to make contact.
(a) Show that the emf induced in the coil is
given (as a function of time) by
  (2 f ) NabB sin(2 ft )   o sin(2 ft ), where 2 f  .
Here, O  2 fNabB is the peak value (or, amplitude) of the emf.
(b) Design a loop that will produce an emf with  0 =150V when rotated at 60.0 rev/s in a
uniform magnetic field of 0.500 T.
Solution:
Eddy currents: In the examples discussed, the currents produced by a changing flux were set up in definite circuits. Often a changing flux sets up circulating currents, called eddy currents, in a piece of bulk metal like the core of a transformer. The heat produced by such currents constitutes a power loss in the transformer. Consider a conducting slab between the pole of an electromagnet. If the field between the poles is changing with time (as it will if the current in the windings is ac), the flux through any closed loop in the slab (such as the flux through the curve C indicated in the figure) will be changing. Since path C is a conducting path, there will be a current along that path. This is just one of many closed paths that will contain currents if B varies. The existence of eddy currents can be demonstrated by pulling copper or aluminum sheet between the poles of a strong magnet. Part of the area enclosed by curve C in the figure is in the field region. As the sheet is pulled to the right, the flux decreases. Assuming the field is into the page, a clockwise current is 
induced (by Lenz law). The magnetic field will exert a left‐directed force ( Fm ) on the loop, opposing the rightward motion of the sheet. Eddy currents are usually unwanted because power is lost as joule heat, and the heat itself must be dissipated. The power loss can be reduced by increasing the resistance of the possible paths for the eddy currents, as shown in Fig (a). Here, the conducting slab is laminated, that is, made up of small strips glued together. Because insulating glue separates the strips, the eddy currents are essentially confined to the strips. The larger eddy‐current loops are broken, and the power loss is greatly reduced. Another way to inhibit eddy currents is to cut slots into the metal sheet, as shown in Fig (b). Eddy currents are not always undesirable. For example, they are often used to damp unwanted oscillations. They are also used in the magnetic braking systems of some rapid transit cars. Additional Problems: 1. A uniform magnetic field of magnitude 2000 G is
parallel to the x axis. A square coil of side 5 cm makes an
angle  with the z axis as shown on the right. Find the
magnetic flux through the coil when (a)   0 , (b)
  30 , (c)   60 , and (d)   90 .
2. The magnetic flux through the loop shown in the figure increases according to
  6.0t 2  7.0t , where  is in milliwebers (mWb) and t in seconds. (a) What is the
magnitude of the emf induced in the loop when t=2.0 s? (b) If R  3.1  , what are the
magnitude and direction of the induced current at
t  2.0 s?
Solution:
(b) The flux is increasing and the induced field opposes this increase, and hence is
directed into the page (  ). To produce this field, the current must flow CW. The
magnitude of the current is

I   0.01 A.
R
The power developed can be found from either P   2 / R or P  I 2 R .
3. The figure on the right shows
four wire loops, with edge lengths of
either L or 2L. All four loops will
move through a region of uniform

magnetic field B (directed out of the page) at the same constant velocity. Rank the four
loops according to the maximum magnitude of the emf induced as they enter the field
region, greatest first.
4. The figure on the right shows an ac
generator. It consists of a rectangular loop
dimensions a and b with N turns
connected to slip rings. The loop rotates
with an angular speed  in a uniform

field B . (a) Show that the potential
difference between the slip rings is
  NBab sin t . (b) If a  1.0 cm,
b  2.0 cm, N  1000 , and B  2 T, at what angular speed  must the coil be rotated to
generate an emf whose maximum value is 110 V?
Solution given on next page.
5. Refer to the figure on the right. A 10-cm by
5-cm rectangular loop with resistance 2.5 is
pulled through a region of uniform magnetic
field B  1.7 T with constant speed v  2.4
cm/s. The front of the loop enters the region of
the field at t  0 .
(a) Find: (i) the flux through the loop, (ii) the
induced emf and current as functions of time.
(b) Graph the three quantities as functions of
time.
Solution given on next page.