Ben Gurion University of the Negev
www.bgu.ac.il/atomchip
Physics 2B for Materials and Structural Engineering
Lecturer: Daniel Rohrlich
Teaching Assistants: Oren Rosenblatt, Shay Inbar
Week 8. Faraday’s law – Magnetism in matter • Faraday’s law of
induction • Lenz’s law • “motional emf” • towards Maxwell’s
equations
Source: Halliday, Resnick and Krane, 5th Edition, Chaps. 34-35.
Faraday’s law of induction
In 1831, Michael Faraday (in England) and Joseph Henry (in
the U.S.) independently discovered that a changing magnetic
flux ΦB through a conducting circuit induces a current!
Source: UCSC
Faraday’s law of induction
In 1831, Michael Faraday (in England) and Joseph Henry (in
the U.S.) independently discovered that a changing magnetic
flux ΦB through a conducting circuit induces a current!
The sign of the current depends on the sign of dΦB/dt.
Faraday’s law of induction
In 1831, Michael Faraday (in England) and Joseph Henry (in
the U.S.) independently discovered that a changing magnetic
flux ΦB through a conducting circuit induces a current!
The sign of the current depends on the sign of dΦB/dt.
Faraday’s law of induction
In 1831, Michael Faraday (in England) and Joseph Henry (in
the U.S.) independently discovered that a changing magnetic
flux ΦB through a conducting circuit induces a current!
The sign of the current depends on the sign of dΦB/dt.
Faraday’s law of induction
In fact, the induced “emf” E is directly proportional to dΦB/dt.
What is an “emf”? It is short for “electromotive force”, which
is not the correct term because an “emf” is not a force. It has
units of volts.
An “emf” is like a potential, but here, evidently, the concept of
a potential doesn’t work.
Physically, an “emf” is an electric field that is created in a
conductor. A better version of Faraday’s law is

d B
E(r)  dr  
dt
.
Faraday’s law of induction
Example 1: A conducting circuit wound 200 times has a total
resistance of 2.0 Ω. Each winding is a square of side 18 cm. A
uniform magnetic field B is directed perpendicular to the plane
of the circuit. If the field changes linearly from 0.00 to 0.50 T
in 0.80 s, what is the magnitude of (a) the induced “emf” E in
the circuit (b) the induced electric field E, and (c) the induced
current I?
Faraday’s law of induction
Example 1: A conducting circuit wound 200 times has a total
resistance of 2.0 Ω. Each winding is a square of side 18 cm. A
uniform magnetic field B is directed perpendicular to the plane
of the circuit. If the field changes linearly from 0.00 to 0.50 T
in 0.80 s, what is the magnitude of (a) the induced “emf” E in
the circuit (b) the induced electric field E, and (c) the induced
current I?
Answer: (a) We calculate
dΦB/dt = (dB/dt) (200) (area)= (0.625T/s) (200) (18 cm)2
= 4.05 T · m2/s = 4.05 W/s = 4.05 V = E.
(The weber W = T · m2 is the MKS/SI unit of magnetic flux,
and since T = N /(m/s) · C = V · s/m2, we have W/s = V.)
Faraday’s law of induction
Example 1: A conducting circuit wound 200 times has a total
resistance of 2.0 Ω. Each winding is a square of side 18 cm. A
uniform magnetic field B is directed perpendicular to the plane
of the circuit. If the field changes linearly from 0.00 to 0.50 T
in 0.80 s, what is the magnitude of (a) the induced “emf” E in
the circuit (b) the induced electric field E, and (c) the induced
current I?
Answer: (b) The total length of the wire is (200) (4) (0.18 m) =
144 m. From the “emf” = 4.05 V we infer E = 4.05 V/144 m =
0.028 V/m.
(c) The current is I = V/R = 4.05 V/2.0 Ω = 2.0 A.
Faraday’s law of induction
Example 2: Two bulbs are connected to opposite sides of a
loop of wire, as shown. A decreasing magnetic field (confined
to the circular area shown) induces an “emf” in the loop that
causes the two bulbs to light. What happens to the brightness
of each bulb when the switch is closed?
Bulb 1
Bulb 2
B
Faraday’s law of induction
Answer: Bulb 2 stops glowing, since it is shorted out, and
Bulb 1 glows brighter, since it is the only resistance in the
circuit.
Bulb 1
Bulb 2
B
Faraday’s law of induction
Example 3: The conducting bar at the right is pulled right with
force Fapp at speed v. The resistance R is the only resistance in
the circuit. The magnetic field B is constant and perpendicular
to the plane of the circuit. What is the current I and what is the
power applied?
v
L
R
FB
Fapp
I
x
Faraday’s law of induction
Answer: The flux ΦB is BLx, so dΦB/dt = BLv. Thus the
current is I = (BLv)/R. The force FB equals BIL so the power
applied is FBv = BILv =I2R, i.e. the power applied is the power
lost in “Joule heating” of the resistor.
v
L
R
FB
Fapp
I
x
Lenz’s law
Let’s see if we can understand not only the magnitude but also
the sign of the current induced by a changing magnetic field.
The figure below is taken from Example 3 with one change:
The direction of the induced current I is reversed.
v
L
R
FB
Fapp
I
x
Lenz’s law
But if the direction of I is reversed, then so is the direction of
FB; then the bar accelerates to the right, v increases, I increases,
FB increases further without limit, and energy is not conserved.
v
L
FB
Fapp
R
I
x
Lenz’s law
But if the direction of I is reversed, then so is the direction of
FB; then the bar accelerates to the right, v increases, I increases,
FB increases further without limit, and energy is not conserved.
v
L
FB
R
I
x
Lenz’s law
But if the direction of I is reversed, then so is the direction of
FB; then the bar accelerates to the right, v increases, I increases,
FB increases further without limit, and energy is not conserved.
Consider also the direction of the magnetic flux generated by I.
v
L
FB
R
I
x
Lenz’s law
These considerations lead us to conclude, with H. Lenz, that
the current induced in a loop by a changing magnetic flux must
generate an opposite magnetic flux through the loop.
v
L
R
FB
Fapp
I
x
Lenz’s law
Example 2: The galvanometer indicates a clockwise current
(seen from above). The south pole of the magnet is down. Is
the hand inserting or withdrawing the magnet?
Lenz’s law
Example 2: The galvanometer indicates a clockwise current
(seen from above). The south pole of the magnet is down. Is
the hand inserting or withdrawing the magnet?
Answer: A clockwise current
implies a downward magnetic
flux. So the flux due to the
magnet must be increasing.
The flux from the south pole
of a magnet increases when the
magnet is inserted.
S
N
Lenz’s law
Example 3: A cylindrical magnet of mass M fits neatly into a
very long metal tube with thin steel walls, and slides down it
without friction. The radius of the magnet is r and the strength
of the magnetic field at its top and bottom is B. The magnet
begins falling with acceleration g. (a) Show that the speed of
the magnet approaches a limiting value v. (b) What is the rate
of heat dissipation in the tube, in terms of v and the other data?
Lenz’s law
Answer: (a) The falling magnet induces a circulating current in
the tube. By Lenz’s law, the magnetic field of this current
opposes the falling magnet, until the magnetic force exactly
balances the force of gravity on the magnet, which falls with
constant speed v. (b) Gravity, the only external force on this
system, does work at the rate Mgv. By energy conservation,
this must be the rate of heat dissipation in the tube.
“Motional emf”
A so-called “motional emf” arises when a conductor moves in
a constant magnetic field. Thus the moving bar (below) is an
example of a “motional emf”. But a “motional emf” can arise
also from the Lorentz force without any magnetic induction.
v
L
R
FB
Fapp
I
x
“Motional emf”
Example 1: A conducting strip of length L moves sideways
with constant velocity v through a constant B pointing out of
the screen. What is the potential difference ΔV between the
two ends of the strip?
B
++
L
v
FB
–
–
“Motional emf”
Answer: At equilibrium, the force on charges anywhere in the
strip must vanish, i.e. E = vB as in the Hall effect. The
potential difference is then ΔV = EL = vBL.
B
++
L
v
FB
–
–
“Motional emf”
Example 2: A conducting strip of length L rotates around a
point O with constant angular frequency ω, in a constant B
pointing out of the screen. What is the potential difference
ΔV between the two ends of the strip?
v
r
dr
L
O
“Motional emf”
Answer: An electron in an element dr of the conducting strip
is subject to a centripetal magnetic force evB which must be
balanced by an electric force eE = evB = eωrB. (Note v is not
uniform along the strip.) Integrating E(r)dr along the strip, we
obtain V 
L
0
L
0
E(r )dr  B rdr  BL2 /2 .
v
r
dr
L
O
Towards Maxwell’s equations
The set of four fundamental equations for E and B,
 E  dA   0
q

 B  dr   0 I
 B  dA  0 ,
d
E  dr    B
dt
(Gauss’s law)
(Faraday’s law)
(Ampère’s law)
together with the Lorentz force law FEM = q (E + v × B), sum
up everything we have learned so far about electromagnetism!
Towards Maxwell’s equations
The set of four fundamental equations for E and B,
 E  dA   0
q

 B  dr   0 I
 B  dA  0 ,
d
E  dr    B
dt
(Gauss’s law)
(Faraday’s law)
(Ampère’s law)
are similar to the famous equations named after J. C. Maxwell
– “Maxwell’s equations” – describing all of electromagnetism.
But they are not yet Maxwell’s equations!
Towards Maxwell’s equations
The set of four fundamental equations for E and B,
 E  dA   0
q

 B  dr   0 I
 B  dA  0 ,
d
E  dr    B
dt
(Gauss’s law)
(Faraday’s law)
(Ampère’s law)
include one equation with an error that Maxwell discovered
and corrected. What is the error?