Faraday's Law

dS

B B

ΦB B dS≡ •∫

ε = −ddtBΦ

Global Review• Electrostatics

» motion of “q” in external E-field» E-field generated by Σqi

• Magnetostatics» motion of “q” and “i” in external B-field» B-field generated by “I”

• Electrodynamics» time dependent B-field generates E-field ac

circuits, inductors, transformers, etc» time dependent E-field generates B-field

• electromagnetic radiation - light

Overview of Lecture

• Induction Effects• Faraday’s Law (Lenz’ Law)

• Energy Conservation with induced currents?

• Faraday’s Law in terms of Electric Fields

Text Reference: Chapter 33: 1 - 5

Induction Effects• Bar magnet moves through coil

⇒ Current induced in coil

v

vS N

• Change pole that enters

⇒ Induced current changes sign vN S

N S• Bar magnet stationary inside coil

⇒ No current induced in coil

• Coil moves past fixed bar magnet

⇒ Current induced in coil S N

Induction Effectsfrom Currents

• Switch closed (or opened)

⇒ current induced in coil b

• Steady state current in coil a

⇒ no current induced in coil b

ab

• Conclusion:

A current is induced in a loop when:• there is a change in magnetic field through it• this can happen many different ways

• How can we quantify this?

An example of induction

downwardvelocity

A wire loop falling into an increasing magnetic field

timemagnetic

field

Force acting on moving charges

N N N

Faraday's Law• Define the flux of the magnetic field through an open surface

as: dS

B BΦB B dS≡ •∫

• Faraday's Law:The emf induced in a circuit is determined by the time rate of change of the magnetic flux through that circuit.

The minus sign indicates direction of induced current (given by Lenz's Law).

ε = −ddtBΦ

So what is this emf??

electro-motive force or emftime

A magnetic field, increasing in time, passes through the blue loop

An electric field is generated “ringing” the increasing magnetic field

Ringing electric field will drive currents, just like a voltage difference

Loop integral of E-field is the “emf”

ε = •∫ E dl

Lenz's Law• Lenz's Law:

The induced current will appear in such a direction that itopposes the change in flux that produced it.

• Conservation of energy considerations:Claim: Direction of induced current must be so as to oppose the change; otherwise conservation of energy would be violated. » Why???

• If current reinforced the change, then the change would get bigger and that would in turn induce a larger current which would increase the change, etc..

vB

S Nv

BN S

1

ACT 1• A conducting rectangular loop moves with

constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown.– What is the direction of the induced

current in the loop?1A

X X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X X

v

x

y

(a) ccw (b) cw (c) no induced current• A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown.

• What is the direction of the induced current in the loop?

(a) ccw (b) cw (c) no induced current

1B v

I

x

y

ACT 1• A conducting rectangular loop moves with

constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown.– What is the direction of the induced

current in the loop?

(c) no induced current(a) ccw (b) cw

1A

X X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X X

v

x

y

• There is a non-zero flux ΦB passing through the loop since B is perpendicular to the area of the loop.• Since the velocity of the loop and the magnetic field are CONSTANT, however, this flux DOES NOT CHANGE IN TIME.• Therefore, there is NO emf induced in the loop; NO current will flow!!

ACT 1• A conducting rectangular loop moves with

constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown.– What is the direction of the induced

current in the loop?

yX X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X X

v

x1A

• The flux through this loop DOES change in time since the loop is moving from a region of higher magnetic field to a region of lower field.• Therefore, by Lenz’ Law, an emf will be induced which will oppose the change of flux.• The current i is induced in the clockwise direction to restore the flux.

(a) ccw (b) cw (c) no induced current• A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown.

• What is the direction of the induced current in the loop?1B

(a) ccw (b) cw (c) no induced current

v

I

x

yi

Calculation

• Suppose we pull with velocity v a coil of resistance R through a region of constant magnetic field B. – What will be the induced current?

» What direction?• Lenz’ Law ⇒ clockwise!!

x x x x x xx x x x x xx x x x x xx x x x x x

vw

x

I

xwBΦB =– What is the magnitude?

» Magnetic Flux:

ε = −ddtBΦ» Faraday’s Law:

IR

wBvR

= =εddt

wBdxdt

wBvBΦ = = ⇒∴

Energy Conservation?

x x x x x xx x x x x xx x x x x xx x x x x x

vw

x

I

F

F'

F'

• Energy is dissipated in circuit at rate P'

• The induced current gives rise to a net magnetic force ( ) on the loop which opposes the motion.

FI wBv

R=

F IwB wBvR

wB w B vR

= = ⎛⎝⎜

⎞⎠⎟

=2 2

P Fv w B vR

= =2 2 2

• Agent must exert equal butopposite force to move the loop with velocity v; ∴ agent does work at rate P, where

′ = = ⎛⎝⎜

⎞⎠⎟

=P I R wBvR

R w B vR

22 2 2 2

P = P' !⇒

∆B/∆t → E• Faraday's law ⇒ a changing B induces

an emf which can produce a current in a loop.

• In order for charges to move (i.e., the current) there must be an electric field.

∴ we can state Faraday's law more generally in terms of the E field which is produced by a changing B field.

x x x x x x x x x xx x x x x x x x x xx x x x x x x x x xx x x x x x x x x xx x x x x x x x x x

r

E

E

E

E

B

• Suppose B is increasing into the screen as shown above. An Efield is induced in the direction shown. To move a charge qaround the circle would require an amount of work =

• This work can also be calculated from ε = W/q.

W qE dl= •∫

∆B/∆t → E• Putting these 2 eqns together:

⇒

• Therefore, Faraday's law can be rewritten in terms of the fields as:

x x x x x x x x x xx x x x x x x x x xx x x x x x x x x xx x x x x x x x x xx x x x x x x x x x

r

E

E

E

E

B

W qE dl= •∫ε = •∫ E dl

ε = Wq

E dl ddtB•∫ = −

Φ

3

• Note that for E fields generated by charges at rest (electrostatics) since this would correspond to the potential difference between a point and itself. Consequently, there can be no "potential function" corresponding to these induced E fields.

E dl•∫ = 0

Rate of change of flux through loop

Line integralaround loop

ACT 2• The magnetic field in a region of space of

radius 2R is aligned with the z-direction and changes in time as shown in the plot.– What is sign of the induced emf in a

ring of radius R at time t=t1?

3A

3B– What is the relation between the magnitudes of the induced electric fields ERat radius R and E2R at radius 2R ?

t

Bz

t1

X X X XX X X X X X X

X X X X X X X XX X X X X X X X X

x

y

X X X X X X X X XX X X X X X X XX X X X X X X

X X X X

R

(a) ε < 0( E ccw)

(b) ε = 0 (c) ε > 0( E cw)

(a) E2R = ER (b) E2R = 2ER (c) E2R = 4ER

ACT 2• The magnetic field in a region of space of

radius 2R is aligned with the z-direction and changes in time as shown in the plot.– What is sign of the induced emf in a

ring of radius R at time t=t1?

3A

t

Bz

t1

X X X XX X X X X X X

X X X X X X X XX X X X X X X X X

x

y

X X X X X X X X XX X X X X X X XX X X X X X X

X X X X

R

(a) ε < 0( E ccw)

(b) ε = 0 (c) ε > 0( E cw)

• There will be an induced emf at t=t1because the magnetic field (and therefore the magnetic flux) is changing. It makes NO DIFFERENCE that at t=t1 the magnetic field happens to be equal to ZERO!• The magnetic field is increasing at t=t1(actually at all times shown!) which induces an emf which opposes the corresponding change in flux. ie electric field must be induced in a clockwise sense so that the current it would drive would create a magnetic field in the -z direction.

Note: the signs work this way:• E cw > 0 means that dS points in -z direction.• dB/dt > 0 ⇒ dΦ/dt < 0• Faraday’s Law ⇒ ε = - dΦ/dt > 0

ACT 2• The magnetic field in a region of space of

radius 2R is aligned with the z-direction and changes in time as shown in the plot.– What is sign of the induced emf in a

ring of radius R at time t=t1?

3A

3B– What is the relation between the magnitudes of the induced electric fields ERat radius R and E2R at radius 2R ?

(a) E2R = ER (b) E2R = 2ER (c) E2R = 4ER

t

Bz

t1

X X X XX X X X X X X

X X X X X X X XX X X X X X X X X

x

y

X X X X X X X X XX X X X X X X XX X X X X X X

X X X X

R

(a) ε < 0( E ccw)

(b) ε = 0 (c) ε > 0( E cw)

• The rate of change of the flux is proportional to the area: dt

dBπRdtdΦ 2B −=

• The path integral of the induced electric field is proportional to the radius.

E d l E R•∫ = ( )2πE R∝

DemoE-M Cannon v

• Connect solenoid to a source of alternating voltage.

~

side view

•

•

F

F

•B

B

B

top view2

• The flux through the area ⊥ to axis of solenoid therefore changes in time.

• A conducting ring placed on top of the solenoid will have a current induced in it opposing this change.

• There will then be a force on the ring since it contains a current which is circulating in the presence of a magnetic field.

ACT 3• For this ACT, we will predict the results of

variants of the electromagnetic cannon demo .– Suppose two aluminum rings are used in the

demo; Ring 2 is identical to Ring 1 except that it has a small slit as shown. Let F1 be the force on Ring 1; F2 be the force on Ring 2.

2B – Suppose two identically shaped rings are used in the demo. Ring 1 is made of copper (resistivity = 1.7X10-8 Ω-m); Ring 2 is made of aluminum (resistivity = 2.8X10-8 Ω-m). Let F1 be the force on Ring 1; F2 be the force on Ring 2.

(a) F2 < F1 (b) F2 = F1 (c) F2 > F1

2ARing 1

Ring 2

(a) F2 < F1 (b) F2 = F1 (c) F2 > F1

ACT 3Ring 1

Ring 2

• For this ACT, we will predict the results of variants of the electromagnetic cannon demo which you just observed.– Suppose two aluminum rings are used in the

demo; Ring 2 is identical to Ring 1 except that it has a small slit as shown. Let F1 be the force on Ring 1; F2 be the force on Ring 2.

2A

(a) F2 < F1 (c) F2 > F1(b) F2 = F1

• The key here is to realize exactly how the force on the ring is produced.• A force is exerted on the ring because a current is flowing in the ring and the ring is located in a magnetic field with a component perpendicular to the current.• An emf is induced in Ring 2 equal to that of Ring 1, but NO CURRENTis induced in Ring 2 because of the slit!• Therefore, there is NO force on Ring 2!

ACT 3Ring 1

Ring 2

• For this ACT, we will predict the results of variants of the electromagnetic cannon demo which you just observed.– Suppose two aluminum rings are used in the

demo; Ring 2 is identical to Ring 1 except that it has a small slit as shown. Let F1 be the force on Ring 1; F2 be the force on Ring 2.

2A

2B – Suppose two identically shaped rings are used in the demo. Ring 1 is made of copper (resistivity = 1.7X10-8 Ω-m); Ring 2 is made of aluminum (resistivity = 2.8X10-8 Ω-m). Let F1 be the force on Ring 1; F2 be the force on Ring 2.

(a) F2 < F1 (c) F2 > F1(b) F2 = F1

(a) F2 < F1 (b) F2 = F1 (c) F2 > F1• The emf’s induced in each ring are equal.• The currents induced in each ring are NOT equal because of the different resistivities.• The copper ring will have a larger current induced (smaller resistance) and therefore will experience a larger force (F proportional to current).