Ben Gurion University of the Negevwww.bgu.ac.il/atomchip

Week 8. Faraday’s law – Magnetism in matter • Faraday’s law of induction • Lenz’s law • “motional emf” • towards Maxwell’s equationsSource: Halliday, Resnick and Krane, 5th Edition, Chaps. 34-35.

Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shay Inbar

Physics 2B for Materials and Structural EngineeringPhysics 2B for Materials and Structural Engineering

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

. )(dt

dd B

rrE

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 (confinedto 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.

B

Bulb 2

Bulb 1

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?

L

x

R

I

vFB

Fapp

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.

L

x

R

I

vFB

Fapp

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.

L

x

R

I

vFB

Fapp

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.

L

x

R

I

vFB

Fapp

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.

L

x

R

I

vFB

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.

L

x

R

I

vFB

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.

L

x

R

I

vFB

Fapp

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 currentimplies a downward magneticflux. So the flux due to themagnet 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.

L

x

R

I

vFB

Fapp

“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?

L

B

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.

L

B

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?

L

O

r

v

dr

“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

L

O

r

v

dr

. /2)( 2

00BLrdrBdrrEV

LL

Towards Maxwell’s equations

The set of four fundamental equations for E and B,

together with the Lorentz force law FEM = q (E + v × B), sum

up everything we have learned so far about electromagnetism!

, 0

0

0

AB

rB

rE

AE

d

Id

dt

dd

qd

B

(Ampère’s law)

(Gauss’s law)

(Faraday’s law)

Towards Maxwell’s equations

The set of four fundamental equations for E and B,

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!

, 0

0

0

AB

rB

rE

AE

d

Id

dt

dd

qd

B

(Ampère’s law)

(Gauss’s law)

(Faraday’s law)

Towards Maxwell’s equations

The set of four fundamental equations for E and B,

include one equation with an error that Maxwell discovered and corrected. What is the error?

, 0

0

0

AB

rB

rE

AE

d

Id

dt

dd

qd

B

(Ampère’s law)

(Gauss’s law)

(Faraday’s law)