Faraday’s Law - University Of IllinoisFaraday’s Discovery True no matter how we change the flux. In fact, the circuit may be stationary (no motional EMF) and only the B-field changes

Physics 212 Lecture 17, Slide 1

Physics 212 Lecture 17

Faraday’s Law


Motional EMF

Change Area of loop Change magnetic field

through loop Change orientation of loop relative to B

B B dA

In each case the flux of the magnetic field through the circuit changes with time and an EMF is produced.

EMF Bd

dt


B

Rotate the loop,change flux, generate emf.


Move loop to a place wherethe B field is different, change flux, generate emf.

B2

v

B1

Checkpoint 1a


• The flux is NOT changing • B does not change • the area does not change • the orientation of B and A does not change

• Motional emf is ZERO • v X B = 0 • no charge separation • no E field • no emf

A copper loop is placed in a uniform magnetic field as shown. You are looking from the right.

Suppose the loop is moving to the right. The current induced in the loop is:

A. zero B. clockwise C. counterclockwise


Current changes direction every time the loop becomes perpendicular with the B field

emf ~ d/dt

(B dA = max) d/dt (B dA ) = 0

X

O B dA X

O

B dA

Checkpoint 1c Now suppose that the loop is spun around a vertical axis as shown, and that it makes one

complete revolution every second.

The current induced in the loop:

A. Is zero

B. Changes direction once per second

C. Changes direction twice per second

Bd

emfdt

Faraday’s Discovery

True no matter how we change the flux. In fact, the circuit may be stationary (no motional EMF) and only the B-field changes with time. An EMF is still produced. This implies that:

Bdemf E dl

dt

Faraday’s Law


B(t) decreasing

Change the B field in time so flux changes. Induce an emf nnd therefore an Electric field. This emf tries to oppose the change in flux. (Lenz’s Law)

dt

ddEemf B

Induces an E field even if there is no circuit there!


Checkpoint 1b

• Motional emf is ZERO • Circuit is stationary !

• HOWEVER: The flux is changing • B decreases in time • current induced to oppose the flux change

• clockwise current tries to restore B that was removed

X X X X X X X X

X X X X X X X X

X X X X X X X X

X X X X X X X X

X X X X X X X X

Looking from right

Clockwise current tries to restore B

Checkpoint 1b A copper loop is placed in a uniform magnetic field as shown. You are looking from the right.

Now suppose the that loop is stationary and that the magnetic field is

decreasing in time. The current induced in the loop is:

A. zero B. clockwise C. counterclockwise


(copper is not ferromagnetic)

This one is hard ! B field increases upward as loop falls

Clockwise current (viewed from top) is induced

Ftotal < mg

a < g

X

O

B

B

Like poles repel

F

Checkpoint 2 A horizontal copper ring is dropped from rest directly above the north pole of a permanent magnet

Will the acceleration a of the falling ring in the presence of the magnet

be any different than it would have been under the influence of just

gravity (i.e. g)?

A. a > g B. a = g C. a < g


Main Field produces horizontal forces “Fringe” Field produces vertical force

I

Looking down

I

B B

IL X B points UP

Ftotal < mg

a < g

HOW IT

WORKS

(copper is not ferromagnetic)

This one is hard ! B field increases upward as loop falls

Clockwise current (viewed from top) is induced

Checkpoint 2 A horizontal copper ring is dropped from rest directly above the north pole of a permanent magnet

Will the acceleration a of the falling ring in the presence of the magnet

be any different than it would have been under the influence of just

gravity (i.e. g)?

A. a > g B. a = g C. a < g


Calculation

• Conceptual Analysis – Once loop enters B field region, flux will be changing in time – Faraday’s Law then says emf will be induced

• Strategic Analysis – Find the emf – Find the current in the loop – Find the force on the current

y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

B A rectangular loop (height = a, length = b, resistance = R, mass = m) coasts with a constant velocity v0 in + x direction as shown. At t =0, the loop enters a region of constant magnetic field B directed in the –z direction. What is the direction and the magnitude of the force on the loop when half of it is in the field?


What is the magnitude of the emf induced in the loop just after it enters the field?

(A) e = Babv02 (B) e = ½ Bav0 (C) e = ½ Bbv0 (D) e = Bav0 (E) e = Bbv0

In a time dt it moves by v0dt

Change in Flux = dB = BdA = Bav0dt

Calculation A rectangular loop (height = a, length = b, resistance = R, mass = m) coasts with a constant velocity v0 in + x direction as shown. At t =0, the loop enters a region of constant magnetic field B directed in the –z direction.

y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

B

y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

B

The area in field changes by dA = v0dt a

a

dt

demf B

oB Bav

dt

d


What is the direction of the current induced in the loop just after it enters the field?

(A) clockwise (B) counterclockwise (C) no current is induced

Flux is increasing into the screen

emf is induced in direction to oppose the change in flux that produced it

Induced emf produces flux out of screen

Calculation y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

B A rectangular loop (height = a, length = b, resistance = R, mass = m) coasts with a constant velocity v0 in + x direction as shown. At t =0, the loop enters a region of constant magnetic field B directed in the –z direction.

y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

B

dt

demf B


What is the direction of the net force on the loop just after it enters the field?

(A) +y (B) -y (C) +x (D) -x

x

y

v0 a

b x x x x x x x

x x x x x x x

B

I • Force on top and bottom segments cancel (red arrows)

Calculation y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x


• Force on right segment is directed in –x direction.

dt

demf B

Force on a current in a magnetic field: BLIF

Physics 212 Lecture 17, Slide 16 x

y

v0 a

b x x x x x x x

x x x x x x x

B

I F

What is the magnitude of the net force on the loop just after it enters the field?

(A) (B) (C) (D)

e = Bav0

R

vaBaB

R

BavF oo

22

Calculation y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x


ILB

F IL B

dt

demf B

RaBvF o4 RBvaF o2 RvBaF o /222 RvBaF o /22

BLIF

since ILBF BL

R

Bav

RI o

e


Follow-Up A rectangular loop (sides = a,b, resistance = R, mass = m) coasts with a constant velocity v0 in + x direction as shown. At t =0, the loop enters a region of constant magnetic field B directed in the –z direction. What is the velocity of the loop when half of it is in the field?

y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

B

t = dt: e = Bav0

Which of these plots best represents the velocity as a function of time as the loop moves form entering the field to halfway through ? (A) (B) (C) D) (E)

This is not obvious, but we know v must decrease Why?

v0 a

b x x x x x x x

x x x x x x x

B

I Fright

Fright points to left Acceleration negative Speed must decrease

X X X


Follow-Up

2 2 /dv

F a B v R mdt

A rectangular loop (sides = a,b, resistance = R, mass = m) coasts with a constant velocity v0 in + x direction as shown. At t =0, the loop enters a region of constant magnetic field B directed in the –z direction. What is the velocity of the loop when half of it is in the field?

Which of these plots best represents the velocity as a function of time as the loop moves form entering the field to halfway through ?

y

x

v0

a

b x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

B

e = Bav0

(A) (D) • Why (D), not (A)? – F is not constant, depends on v

Challenge: Look at energy Claim: The decrease in kinetic energy of loop is equal to the energy dissipated as heat in the resistor. Can you verify??

dt

dvm

R

vBaF

22

where

toevv

mR

Ba 22

Documents

Faraday’s Law - University Of IllinoisFaraday’s Discovery True no matter how we change the flux. In fact, the circuit may be stationary (no motional EMF) and only the B-field changes