Lecture 10 Induction and Inductance Ch. 30• Cartoon - Faraday Induction• Opening Demo - Thrust bar magnet through coil and measure the current• Topics

– Faraday’s Law– Lenz’s Law– Motional Emf– Eddy Currents– LR Circuits– Self and mutual induction

• Demos Polling

Demos1. Galvanometer with bar magnet and 3 coils.

2. 2. Gray magnet, solenoid, two LED’s, push and pull, shows different LED’s light up. Lenz’s Law.

3. Coil connected to AC source will induce current to light up bulb in second coil.

4. Jumping aluminum rings from core of solenoid powered by an AC source. Cool a ring.

5. Slowing down of swinging copper pendulum between poles faces of a magnet. Slitted copper pendulum. Eddy Currents

6. Inductive spark after turning off electromagnet. Inductance.

7. Neodymium magnet swinging over copper strip. Eddy currents

8. Neodymium magnet falling through copper pipe. Then cool with liquid nitrogen. Eddy currents

9. Hanging aluminum ring with gray magnet. Lenz’s Law

10. Two large copper disks with one magnet levitating above another

Introduction Stationary charges cause electric fields

(Coulombs Law, Gauss’ Law).

Moving charges or currents cause magnetic fields (Biot-Savart Law).

Therefore, electric fields produce magnetic fields.

Question: Can changing magnetic fields cause electric fields?

• Discovered in 1830s by Michael Faraday and Joseph Henry. Faraday was a poor boy and worked as a lab assistant and eventually took over the laboratory from his boss.

• Faraday’s Law says that when magnetic flux changes in time, an Emf is induced in the environment which is not localized and also is non-conservative.

• Lets look at various ways we can change the magnetic field with time and induce an Emf. If a conductor is present, a current can be induced.

Faraday's Law Emf =−dφmdt

1. What is magnetic flux ?φm

2. What is an induced Emf ?

Magnetic flux

∫ ⋅= dAnB ˆr

φ

Field from Bar magnet

Faraday’s Law

€

Emf = −dφ

dt

Demo 1 Thrust a bar magnet through a loop of wire

Current flows in the ring in a direction that produces a magnetic field that oppose the bar magnet.

Units:

B is in Tesla

A is in m2

is in (Webers) Wbφm

A

n̂ i

R

iB

20μ= at center

Produced by current flowing in the wire.

The current flows in the wire to produce a magnetic field that opposes the bar magnet. Note North poles repel each other.

Lenz’s Law

Lenz’s Law

B produced by bar magnet

More on Lenz’s Law: An induced current has a direction such that the magnetic field

due to the current opposes the change in themagnetic flux that induces the current

Question: What is the direction of the current induced in the ring given B increasing or decreasing?

B due to induced current

B due to induced current

Demo 2: Gray magnet, solenoid, Red and Green LEDs

Push gray magnet in, B field points to the right . Current starts to flow in coil to produce a field to oppose the initial field. Red LED lights up.

Pull magnet out, the opposite happens and the Green LED lights up.

N

magnet solenoid

repel

B

N

iGreen

Red

Demo 3: Coil connected to AC sourceLight bulb connected to second coil

Shows how flux changing through one coil due to alternating current induces current in second coil to light up bulb. Note no mechanical motion here.

Coil 1

Coil 2Iron core (soft)

means lots of inductance in wire so AC doesn’t heat up wire.

Demo 4: Jumping aluminum ring from core of solenoid powered by an AC source.

• When I turn on the current, B is directed upward and momentarily the top of the iron is the North pole. If the ring surrounds the iron, then the flux in it increases in the upward direction. This change in flux increases a current in the ring so as to cause a downward B field opposing that due to the solenoid and iron. This means the ring acts like a magnet with a North pole downward and is repelled from the fixed coil.

• Try a square-shaped conductor• Try a ring with a gap in it• Try a ring cooled down to 78 K

Coil

Iron core (soft)

N

AC source

induced current induced B field

Repulsive force because 2 North poles

N

Pull a conducting bar in a magnetic field. What happens to the free charges in the material? Moving bar of length L and width W entirely immersed in a magnetic field B. In this case an Emf is produced but no current flows

€

×

€

×

€

×

€

×

- vF

+ vF

L

B

Positive charges pile up at the top and negative charges at the bottom and no current flows, but an Emf is produced. Now let’s complete the circuit.

W

Motional Emf- field fixed but conductor moves

qvBLemfqUqvBLWork

=×==

BqvFm ×=

€

emf = −dφmdt

= −BdA

dt

Pull the rectangular loop out of the magnetic field. A current i will be induced to flow in the loop in the direction shown. It produces a magnetic field that tries to increase the flux through the loop.

Motional EmfWhat force is required to keep current flowing in the circuit?

Uniform magnetic field into the screen

A= area of magnetic fieldenclosed by the wire

FA

Wire

+ vClose up of the wire

BqvFm ×=

BiLF ×=1

Faraday and Lenz's Law

emf =−dφm

dr=−B

dAdt

=−Bd(Lx)dt

=−BLdxdt

=−BLv

cancel

R is the resistance

Uniform magnetic field into the screen

€

F1 =B2L2v

R

This is the force you need to pull at to achieve constant speed v.

Motional Emf Continued

F1=FA

FA

F1 =iL×B Lorentz Law

23

2

1

FFBixFBiLF

−===

emf =iR Ohms LawBLv=iR

i =BLvR

Motional Emf : Work Done by me

How much work am I doing in pulling the circuit?

What is the rate at which I am doing work? P=Fv

€

P = F1v =B2L2v 2

RWhat is the thermal energy dissipated in the loop?

€

P = i2R

i =BLv

R

P =B2L2v 2

RNote that the rate at which I do work in pulling the loop appears totally as thermal energy.

Circuit diagram for motional Emf. R is the resistance of the wire

(Note that the magnetic field does not do any work,)

W =F1d=B2L2vdR

Eddy Currents

A solid piece of copper is moving out of a magnetic field. While it is moving out, an emf is generated forming millions of current loops as shown.

Eddy currents are also formed in a copper pendulum allowed to swing across a magnet gap cutting magnetic lines of flux. Note that when the copper plate is immersed entirely in the magnet no eddy currents form.

.

• Demo 5: Show a copper plate swinging through a magnet.

• For example, in pulling it out, that part of the plate that was in the B field experiences a decrease in B and hence a change in magnetic flux in any loop drawn in that part of the copper. An emf is developed around such loops by Faraday's Law and in such a direction so as to oppose the change.

B inducedB

SHorseshoe magnet

Copper penduluminduced current

N

Pull back pendulum and release. Pendulum dampens quickly. Force acts to slow down the pendulum.

• Demo 5 Cont. Also try copper plate with slits. What happens now?

• Demo 5 Cont. Note inductive spark when turning power off

• Demo 6: Show neodymium magnet swinging over

copper strip.

Demo 7: Copper pipe and neodymium-iron-boron magnet with magnetic dipole moment

Copper pipe

S

N

W.M. Saslow Am. J. Phys. 60(8)1977

NN

iN

FD

ad

B

iB

Two Norths repel so the magnet

drops more slowly.

Cool down the copper pipe with liquid nitrogen 28 K. What do you expect to happen?

FD=Magnetic Drag Force ~σμ2d

a4

σ is the conductivity of copper

σ =1

ρ

€

σ =5.9 ×106 at 293 K

σ = 149 ×106 at 78 K

Ratio = 25 times more conductive at 78 K

Demo 8: Hanging aluminum ring with gray magnet

• Move magnet toward ring – they repel

Current induced in ring so that the B field produced by the current in the ring opposes original B field.

This means the ring current produces a N pole to push away the N pole of the permanent magnet.

• When magnet is pulled back, it attracts the ring.

i

N NB

repel

induced current

induced B field

Note 2 N poles.

i

N SB

NCurrent in ring is opposite to that above

The orange represents a magnetic field pointing into the screen and let say it is increasing at a steady rate like 100 gauss per sec. Then we put a copper ringIn the field as shown below. What does Faradays Law say will happen?

Current will flow in the ring. What willhappen If there is no ring present?

Now consider ahypothetical pathWithout any copper ring.There will be an induced Emf withelectric field lines asshown above.

In fact therewill be many concentriccircles everywhere in space.

The red circuitshave equal areas.Emf is the same in 1 and 2, less in 3and 0 in 4. Note no current flows. Therefore, no thermal energy is dissipated

Changing magnetic field generates electric field

We can now say that a changing magnetic field produces an electric field not just an Emf. For example:

Work done in moving a test charge around the loop in one revolution of induced Emf is

Work done is also )2(00 rEqdsEqdsF π∫∫ =⋅=⋅

Hence, Emf = 2πrE or more generally for any path

∫ ⋅= dsEEmfdt

ddsE BΦ

−=⋅∫ Faraday’s Law rewritten

But we can not say ∫ ⋅−=−f

iif dsEVV because it would be 0.

Electric potential has no meaning for induced electric fields

0qemfWork ×=

SummaryCharacteristics of the induced emf

• The induced emf is not localized such as at the terminals of a battery.

• It is distributed throughout the circuit.• It can be thought of as an electric field circulating around a

circuit such that the line integral of the electric field taken around a closed loop is the emf.

• Since the line integral is not 0, the field is non-conservative.• There are no equipotential surfaces.• If there is a conductor present, a current will flow in the

conductor.• If no conductor is present, there is no current flow, only emf.• Energy is dissipated only if charges are present.

Example: A magnetic field is perpendicular to the board (screen) and uniform inside a radius R. What is the magnitude of the induced field at a distance r from the center?

Notice that there is no wire or loop of wire. To find E use Faraday’s Law.

dt

drEEdl mφπ −==∫ 2

2rBBAm πφ ==

dt

dBrrB

dt

d

dt

d m 22 )( ππφ==

dt

dBrrE 22 ππ −=

Rrdt

dB

r

RE >−=

2

2

Rrdt

dBrE <−=

2

E is parallel to dl

R

x xx

x

x

xxx

x

x r

ldr

Field circulates around B field

B

dt

dBRrE 22 ππ =

Example with numbers

Suppose dB/dt = - 1300 Gauss per sec and R= 8.5 cm

Rrdt

dBrE <−=

2

Find E at r = 5.2 cm

Find E at 12.5 cm

Rrdt

dB

r

RE >−=

2

2

mmV

mVT

mm

E 8.30038.013.0)125.0(2

)085.0( 2

===

mmVT

mE 4.30034.013.0

2)052.0(

mV ===

What is an inductor?

Suppose I move the switch to position a, then current starts to increase through the coil. An Emf is induced to make current flow in the opposite direction.

An inductor is a piece of wire twisted into a coil. It is also called a solenoid. If the current is constant in time, the inductor behaves like a wire with resistance. The current has to vary with time to make it behave as an inductor. When the current varies the magnetic field or flux varies with time inducing an Emf in the coil in a direction that opposes the original change.

Now suppose I move the switch to position b

What is inductance?

(H=Henry)

(Henry/m)

1 H=1 T.m2/A

A

i B

l

n̂ N turns

€

L = μ0n2lA

€

L

l= μ0n

2A

μ0 = 4π × 10−7 T ⋅m/A

€

L =NΦ

i

L =(nl)(BA)

i=

(nl)(μ 0ni)(A)

iB = μ 0nl

L = μ 0n2lA

What is inductance? L= Nϕ/i

Emf =−dΦdt

Φ =NBA

=−NAdBdt

B=μ0ni for a solenoid

=−μ0nNAdidt

N=nl

Emf =−μ0n2lA

didt

(H=Henry)

(Henry/m)

1 H=1 T.m2/A

A

i B

l

n̂ N turns

Faraday’s Law

€

L = μ0n2lA

€

L

l= μ0n

2A

μ0 = 4π × 10−7 T ⋅m/A

Emf =−Ldidt

Area

Show demo: Inductive spark after turning off electromagnet

Numerical example – how many turns do you need to make a L = 4.25 mh solenoid with l = 15 cm and radius r = 2.25 cm?

TurnsA

LlnlN 565)0225(.104

)15)(.00425(.27

0

=×

=== − ππμ

A

i B

l

n̂ N turns

€

L = μ0n2lA n =

Lμ0lA

N =nl

Numerical Example

You have a 100 turn coil with radius 5 cm with a resistance of 10 Ω. At what rate must a perpendicular B field change to produce a current of 4 A in the coil?

Emf = IR = (4A)(10Ω) = 40 Volts

€

Emf = −Ndφmdt

= −NAdB

dt= −Nπr2 dB

dt= 40

sT

mV

rNdtdB 51

)05(.15.31004040

22=

⋅⋅==

π

Multiply by N

B(t)

N

N coils sodt

dNemf mφ−=

N = 100 turns

R = 5 cm

Coil resistance = 10

€

ε−iR − Ldi

dt= 0

0=++ LR VVε

RL Circuits

Loop Rule: Sum of potentials =0

€

VR = Ri

€

VL

The potential can be defined across the inductor outside the region where the magnetic flux is changing.

Close the switch to a.

What happens? Write down the loop rule.

Solve this equation for the current i.

Ri

)1( L

Rt

R eV−

−=ε

L

Rt

L eV−

=ε

dt

diL

L

R=τ

τ

RRi

)1( L

Rt

R eV−

−=ε

L

Rt

L eV−

=ε

dt

diL

€

τ = L

R

τ

R€

i =ε

R(1− e

−Rt

L )

€

VR = Ri

Note τ = L/R = 4/2000 = 0.002 s,

€

VR = ε(1− e−1) = 0.63ε

and

€

i =ε

R(1− e−1) = 0.63

ε

R

€

VL = ε(e−1) = 0.37ε

€

VR = 0.63ε

2000

10 V4.0H

How is the magnetic energy stored in a solenoid or coil in our circuit?

€

ε−iR − Ldi

dt= 0

€

ε =iR + Ldi

dt

€

εi = i2R + Lidi

dt

Rate at which energy is delivered to circuit from the battery

Rate at which energy is lost in resistor

Rate at which energy isstored in the magnetic fieldof the coil

Start with Loop rule or Kirchhoff's Law I

€

dUB

dt= Li

di

dt

Solve it for ε

Multiply by i

What is the magnetic energy stored in a solenoid or coil

€

dUB

dt= Li

di

dt

€

dUB = Lidi

€

dUB0

UB∫ = Lidi0

i

∫

€

UB = Lidi0

i

∫ = 12 Li

2

€

UB = 12 Li

2For an inductor L Now define the energy per unit volume

€

uB =UB

Al

Area A l

€

uB =12 Li

2

Al=L

l

i2

2A

€

L

l= μ0n

2A

€

uB =12 Li

2

Al= 1

2 μ0n2i2

€

uB =B2

2μ0 €

B = μ0ni

€

uE =E 2

2ε0

The energy density formula is valid in general

What is Mutual Inductance? M

When two circuits are near one another and both have currents changing, they can induce emfs in each other.

On circuit boards you have to be careful you do not put circuits near each other that have large mutual inductance.

They have to be oriented carefully and even shielded.

221111 IMILm +=φ

112222 IMILm +=φ

MMM == 2112

1 2I1

I2

Chapter 30 Problem 15

In Figure 30-44, a stiff wire bent into a semicircle of radius a is rotated at constant frequency f in a uniform magnetic field B. (Use a, f and B as necessary, with all quantities in SI units.)

(a) What is the frequency of the emf induced in the loop?(b) What is its amplitude?

Chapter 30 Problem 34

Figure 30-56 shows two circular regions R1 and R2 with radii r1 = 12.0 cm and r2 = 34.0 cm. In R1 there is a uniform magnetic field of magnitude B1 = 50.0 mT into the page and in R2 there is a uniform magnetic field B2 = 75.0 mT out of the page (ignore fringing). Both fields are decreasing at the rate of 7.80 mT/s.

• Calculate the integral line int E·ds • for each of the three dashed paths.• (a) path 1• (b) path 2• (c) path 3

Chapter 30 Problem 35

A long solenoid has a diameter of 12.0 cm. When a current i exists in its windings, a uniform magnetic field B = 27.0 mT is produced in its interior. By decreasing i, the field is caused to decrease at the rate of 7.90 mT/s. Calculate the magnitude of the induced electric field at the following distances from the axis of the solenoid.

(a) 2.20 cm(b) 8.20 cm