Electromagnetic InductionElectromagnetic Induction

Chapter 31Chapter 31(cont.)(cont.)

Faraday’s Law & Lenz’s LawFaraday’s Law & Lenz’s LawInduced ElectricInduced Electric FieldsFields

Mutual InductanceMutual InductanceSelf-InductanceSelf-Inductance

Inductors in CircuitsInductors in CircuitsMagnetic EnergyMagnetic Energy

Electromagnetic InductionFaraday’s Law

A changing magnetic field induces an emfproportional to the rate of change of magnetic flux:

= - dB / dt

The emf is about some loop – and the flux is through that loop.

€

B =r B ⋅d

r A ∫ = Bcosθ dA∫

Lenz’s law: the induced emf is directed so that any induced current opposes the change in magnetic flux that causes the

induced emf.

Rotating Loop - The Electric Generator

Consider a loop of area A in a uniform magnetic field B.Rotate the loop with an angular frequency .

A

B

The flux changes because angle changes with time: = t. Hence

dB/dt = d(B · A)/dt

= d(BA cos )/dt

= BA d(cos(t))/dt

= - BA sin(t)

A

• Then by Faraday’s Law this motion causes an emf

= - dB /dt = BA sin(t)

• This is an AC (alternating current) generator.

BA

dB/dt = - BA sin(t)

Rotating Loop - The Electric Generator

Consider a stationary wire in a time-varying magnetic field. A current starts to flow.

Induced Electric Fields

x dB/dt

So the electrons must feel a force F.

It is not F = qvxB, because the charges started stationary.Instead it must be the force F=qE due to an

induced electric field E.

That is:A time-varying magnetic field B

causes an electric field E to appear!

Consider a stationary wire in a time-varying magnetic field. A current starts to flow.

Induced Electric Fields

x dB/dt

So the electrons must feel a force F.

It is not F = qvxB, because the charges started stationary.Instead it must be the force F=qE due to an

induced electric field E.

Moreover E around the loop gives a voltage diff V=E·dl.And this must be the emf:

= E·dlo

Induced Electric Fields

E·dl = - dB/dto

This gives another way to write Faraday’s Law:

A technical detail:

The electrostatic field E is conservative: E·dl = 0.Consequently we can write E = - V.

The induced electric field E is NOT a conservative field.We can NOT write E = -V for an induced field.

o

Electrostatic Field Induced Electric Field

F = q E F = q E

Vab = - E·dl

E·dl = 0 and Ee = V

E · dl = - dB/dt

E·dl 0

Conservative Nonconservative

Work or energy difference does NOT depend on path

Work or energy difference DOES depend on path

Caused by stationarycharges

Caused by changingmagnetic fields

o

o

o

Induced Electric Fields

x B E · dl = - dB/dto

Faraday’s Law

Now suppose there is no conductor:Is there still an electric field?

YES! The field does not depend on the presence of the conductor.

For a dB/dt with axial or cylindrical symmetry, the field lines of E are circles. dB/dt

E

Mutual Inductance

• Two coils, 1 & 2, are arranged such that flux from one passes through the other.

• We already know that changing the current in 1 changes the flux (in the other) and so induces an emf in 2.

• This is known as mutual inductance.

IBof 1 through 2

1 2

Mutual Inductance

The mutual inductance M is the proportionality constant between 2 and I1:

2 = M I1

so d2 /dt = M dI1 /dtThen by Faraday’s law:

2= - d2 /dt = - M dI1 /dt

Hence M is also the proportionality constant

between 2 and dI1 /dt.

Bof 1 through 2I

1 2

Mutual Inductance

• M arises from the way flux from one coil passes through the other: that is from the geometry and arrangement of the coils.

• Mutual means mutual. Note there is no subscript on M: the effect of 2 on 1 is identical to the effect of 1 on 2.

• The unit of inductance is the Henry (H).

1 H = 1Weber/Amp = 1 V-s/A

• If the current is steady, the coil acts like an ordinary piece of wire.

• But if the current changes, B changes and so then does , and Faraday tells us there will be an induced emf.

• Lenz’s law tells us that the induced emf must be in such a direction as to produce a current which makes a magnetic field opposing the change.

Self Inductance

I B

A changing current in acoil can induce an emfin itself….

Self Inductance

• The self inductance of a circuit element (a coil, wire, resistor or whatever) is L = B/I.

• Then exactly as with mutual inductance = - L dI/dt.

• Since this emf opposes changes in the current (in the component) it is often called the “back emf”.

• From now on “inductance” means self-inductance.

L = 0n2Ad

What is the (self) inductance of a solenoid with area A, length d, and n turns per unit length?

In the solenoid B = 0nI, so the flux through one turn

is B = BA = 0nIA

The total flux in the solenoid is (nd)B

Therefore, B = 0n2IAd and so L = B/I gives

Example: Finding Inductance

(only geometry)

Inductance Affects Circuits and Stores Energy

• This has important implications…..• First an observation: Since cannot be

infinite neither can dI/dt. Therefore, current cannot change instantaneously.

• We will see that inductance in a circuit affects current in somewhat the same way that capacitance in a circuit affects voltage.

• A ‘thing’ (a component) with inductance in a circuit is called an inductor.

RL Circuit

R

LWhile the switch is open current can’t flow.

We start with a simple circuit containing a battery, a switch, a resistor,

and an inductor, and follow what happens when the switch is closed.

+-

S

0

RL Circuit

R

L

L

When the switch is closed current I flows, growing gradually, and a

‘back emf’ L=- L dI/dt is generated in the inductor. This opposes the current I.

While the switch is open current can’t flow.

We start with a simple circuit containing a battery, a switch, a resistor,

and an inductor, and follow what happens when the switch is closed.

+-

S

I0

+-

S

0

RL Circuit

R

L

L

After a long time the current

becomes steady. Then L is zero.

When the switch is closed current I flows, growing gradually, and a

‘back emf’ L=- L dI/dt is generated in the inductor. This opposes the current I.

While the switch is open current can’t flow.

We start with a simple circuit containing a battery, a switch, a resistor,

and an inductor, and follow what happens when the switch is closed.

0Is

+-

S

I0

+-

S

0

+-

S

0

RL Circuit

0/R

I

00 1 2 3 4 5

t/(L/R)

the current I increases exponentially from I = 0 to I = 0/R

When the switch S is closed

+-

S

I0

RL Circuit Analysis

Use the loop method

0 - IR - LdI/dt = 0

LdI/dt = 0 - IR = -R[I-(0 /R)]

dI / [I-(0 /R)] = - (R/L) dt dI / (I-(0 /R)) = - (L/R) dt

ln[I-(0 /R)] – ln[-(0 /R)] = - t / (L/R) ln[-I/(0 /R) + 1] = - t / (L/R)

-I/(0 /R) + 1 = exp (- t / (L/R)) I/(0 /R) = 1 - exp (- t / (L/R))

I = (0 /R) [1 - exp (- t / (L/R))]

R+-

S

I0 L

RC Circuit Analysis

0 /R

I

00 1 2 3 4 5

t/(L/R)

I = (0/R) [1 - exp (- t / (L/R))]

The current increasesexponentially

with time constant = L / R

t = 0 I = 0

t = I = 0 /R

R+-

S

I0 L

Inductor’s emf L

L

0

0 1 2 3 4 5

t/(L/R)

-0

L = - L (dI/dt)

I = (0/R) [1 - exp (- t / (L/R))]

L = L (0/R) (-R/L) exp (- t / (L/R))

L = - 0 exp (- t / (L/R))

t = 0 L = - 0

t = L = 0

R+-

S

I0 L

Decay of an RL Circuit

• After I reaches 0/R move the switch as shown.

• The loop method gives L - IR = 0 or L = IR

• Remember that L = -L dI/dt -L dI/dt = IR

dI/I = - dt / (L/R) dI/I = - dt / (L/R)• ln I/I0 = - t / (L/R) I = I0 exp [- t / (L/R)]

• But I0 = 0/R

• Then: I = (0/R) exp [- t / (L/R)]

R+-

S

0 L

Inductors in Circuits

• The presence of inductance prevents currents from changing instantaneously.

• The time constant of an RL circuit is = L/R.

• When the current is not flowing the inductor tries to keep it from flowing. When the current is flowing the inductor tries to keep it going.

Energy Stored in an Inductor

Recall the original circuit when current was changing

(building up). The loop method gave: 0 - IR + L = 0

Multiply by I and use L = - L dI/dt

Then:

I0 - I2R - ILdI/dt = 0or:

I0 - I2R – d[(1/2)LI2]/dt = 0 {d[(1/2)LI2]/dt=ILdI/dt}

R+-

S

I0 L

• Think about I0 - I2R - d((1/2)LI2)/dt = 0

• I0 is the power (energy per unit time) delivered by the battery.

• I2R is the power dissipated in the resistor.

• Think about I0 - I2R - d((1/2)LI2)/dt = 0

• I0 is the power (energy per unit time) delivered by the battery.

• I2R is the power dissipated in the resistor.• Hence we’d like to interpret d((1/2)LI2)/dt as the rate at

which energy is stored in the inductor.In creating the magnetic field in the inductor

we are storing energy

UB = (1/2) LI2

• Think about I0 - I2R - d((1/2)LI2)/dt = 0

• I0 is the power (energy per unit time) delivered by the battery.

• I2R is the power dissipated in the resistor.• Hence, we’d like to interpret d[(1/2)LI2]/dt as the rate

at which energy is stored in the inductor.In creating the magnetic field in the inductor

we are storing energy• The amount of energy in the magnetic field is:

Energy Density in a Magnetic Field

• We have shown

• Apply this to a solenoid:

• Dividing by the volume of the solenoid, the stored energy density is:

• This turns out to be the

energy density in a magnetic field

€

UB = 1

2LI2

uB = B2/(20)

€

UB = 1

2μon

2Al I2 =Al

2μo

μo2n2I2

( ) =Al

2μo

B2

Summary

• We defined mutual and self inductance,• Calculated the inductance of a solenoid.• Saw the effect of inductance in RL circuits.• Developed an expression for the stored energy.• Derived an expression for the energy density of a

magnetic field.• Next class we will start learning about alternating-

current (AC) circuits, containing resistors, capacitors, and inductors.