Inductance

Fig. 30.3

What difference does a coil of wire make in

a circuit? In a d.c. circuit, there is no difference

between a coil and a straight wire. But in an a.c.

circuit, a coil of wire makes a big difference, with

changing currents, and induced emfs.

A changing current in one coil can induce an emf

in another coil (as in the figure). The coupling between

two coils is described by mutual inductance.

coil 1

coil 2

PHYS 153 08W 1PHYS 153 08W 1PHYS 153 08W 1

Consider the two coils shown. The black coil is coil 1, and the blue coil is

coil 2.

Denote time varying quantities with lower case letters.

Current creates a magnetic field in coil 1, and some of this flux (or all of it

in this case) passes through coil 2. Denote this flux, per turn, . 1i

2BΦ

12 iB ∝Φ

1PHYS 153 08W

or 12122 iMN B =Φ

21M is the mutual inductance between the two coils.1

2221

i

NM BΦ

=

dt

diM

dt

dN B 1

212

2 =Φ

∴

Since dt

dN B222

Φ−=ε

dt

diM 1

212 −=∴ε

21M depends on the geometry of the two coils.

Starting with coil 2, and considering the interaction with coil 1, we should

use as the mutual inductance. But the interaction between the two 12M

PHYS 153 08W 2PHYS 153 08W 2PHYS 153 08W 2PHYS 153 08W 2

use as the mutual inductance. But the interaction between the two

coils is the same whether you put a changing current through coil 1, or a

changing current through coil 2.

12M

2

11

1

222112

i

N

i

NMMM BB Φ

=Φ

===

Reciprocal relationships hold.

dt

diM 1

2 −=ε and dt

diM 2

1 −=ε

(unit is the henry, H)

1 H = 1Wb/A = 1 V.s/A

If the current in coil 1 varies, there will be an induced emf in coil 1 as well (and

only coil 1 if there is no coil 2). This is a self-induced emf, and by Lenz’s law,

will always oppose the change that caused it.

We define self-inductance by i

NL BΦ= (units are the same, henry, or H)

From this, we can see that

dt

diL

dt

dN B =

Φ

diL−=ε

PHYS 153 08W 3PHYS 153 08W 3PHYS 153 08W 3PHYS 153 08W 3

The induced emf dt

diL−=ε

Note that the equations for self-inductance have the same form as those

for mutual inductance.

Because of this opposition for changes to the current, an inductor (or

choke as it is sometimes called) has resistance in an a.c. circuit. It can

also act with a resistor to cause resonance at a particular frequency in

a circuit. This is the principle behind being able to tune a radio to a

specific frequency, or radio station.

Fig. 30.11

Here is an example of an RL −circuit. When S1 is open, there is no

current. Just after S1 is closed, the

current is still zero. But there is a back

emf in the inductor. A short

time after the switch is closed, there is

a current in the circuit and a potential

dtdiL

iRdrop across the resistor.

Applying Kirchoff’s loop rule, (S2 always open)

PHYS 153 08W 4PHYS 153 08W 4PHYS 153 08W 4

Applying Kirchoff’s loop rule,

0=−−dt

diLiRε

Fig. 30.8

Self-inductance of a toroid.

Assume that B is uniform across the cross-

section.

r

NiABAB π

µ2

0==Φ

The flux is the same through each turn, and there are turns. BΦ N

r

AN

i

NL B

πµ2

2

0=Φ

=∴

For turns, 500=N20.4 cmA = , and cmr 5.7=

HL µ267=

If the current increased from 0 to 4.0 A in 2.0 sµ

||||di

L=ε

PHYS 153 08W 5PHYS 153 08W 5PHYS 153 08W 5

||||dt

diL=ε = 533.3 V opposing the current increase.

Inductance of a solenoid

Assume that is uniform inside and zero outside the solenoid.

The solenoid has turns, cross-sectional area , and length .

→

BN A l

il

NniB 00 µµ ==

l

NiAB

0µ=Φ

l

AN

l

NiA

i

N

i

NL oB

2

0µµ==

Φ=∴

Problem 30.44

A coil has 400 turns and an inductance of 3.50 mH. The current in the coil

varies with time according to

).0250.0

cos()680(s

tmAi π=

(a) What is the maximum emf induced in the coil?

(b) What is the maximum average flux through each turn of the coil?

(c) At t = 0.0180 s, what is the magnitude of the induced emf?

PHYS 153 08W 6PHYS 153 08W 6PHYS 153 08W 6

Energy stored in an inductor

Fig. 24.2

Fig. 28.24

→

B

PHYS 153 08W 7PHYS 153 08W 7PHYS 153 08W 7

Recall that it took work to charge a capacitor. It took no work to place the first

charge, and the most work to place the last charge. What happened to all the

work that was done? It went into establishing an electric field.

The work done, or the energy stored was calculated to be

QVCVC

QU

2

1

2

1

2

22

===

This energy was recovered when the capacitor was discharged.

Similar ideas apply to an inductor as well.

The energy density 2

02

1

.E

volUu ε==

If the current in an inductor, , is increasing at the rate , there will be

opposition to further current flowing into the conductor. i dt

di

Assuming that the resistance of the windings is zero, the voltage between

the terminals a and b is

R

dt

diLVab =

The rate P, at which at which energy is being delivered to the inductor (equal

to the instantaneous power supplied by the external source) is

diLiiVP ==

PHYS 153 08W 8PHYS 153 08W 8PHYS 153 08W 8

dtLiiVP ab ==

During a time , the energy supplied is dt LidiPdtdU ==

Hence the total energy supplied in building up to a final current is

2

0 2

1LIidiLU

I

∫ ==

I

This work done, or energy supplied, goes into establishing a magnetic field.

To develop these ideas further, consider a toroidal solenoid (toroid). Here

the magnetic field is completely contained, with no fringing effects at the end.

The self inductance , and so for a toroid, r

ANL

πµ2

2

0=

r

AINLIU

πµ22

1

2

1 22

02 ==

The magnetic energy density is

2

22

0)2(2

1

2. r

IN

rA

U

vol

Uu

πµ

π===

Recall for a solenoid, r

NIB

πµ2

0=

2B

PHYS 153 08W 9PHYS 153 08W 9PHYS 153 08W 9

0

2

2µB

u =∴ (magnetic energy density in a vacuum)

or µ2

2Bu = (magnetic energy density in some material)

Note the similarity in form to the energy density stored in an electric field.

Even though we relied on the special case of a toroid, this expression is valid

in general.

Fig. 20.5 Fig. 30.10

Recall way back in the study of thermodynamics, --the Otto cycle.

PHYS 153 08W 10PHYS 153 08W 10PHYS 153 08W 10

At the beginning of the adiabatic expansion (the power stroke), the gas was

compressed and the spark plug fired.

There is a primary coil connected to the battery, which produces a strong

magnetic field. A secondary coil with about 100 x as many turns is wound

around the primary. At the instant the plug is required to fire, the current to the

primary is interrupted. This produces an emf of tens of thousands of volts in

the secondary coil, which is connected across the spark plug, causing a spark.

This energy comes from the stored magnetic field.

R-L circuit

We have already had a brief look at this back in slide 4.

Let’s now look at current growth ( charging) and current decay

( discharging).≈

≈A simple circuit with an R and L is

shown.

Why is the switch S2 included?

At instant that S1 is closed, 0=i

PHYS 153 08W 11PHYS 153 08W 11PHYS 153 08W 11

Fig. 30.11

At instant that S1 is closed,

and all of the voltage drop is across

L. When current starts flowing, there

will be a voltage drop across R as

well.

0=i

0=−−dt

diLiRε

L

iR

dt

di −=ε

At ,0=t ,0=i and Ldt

dit

ε==0)(

,∞=t ,Ii = ,0)( =∞=tdt

di

RI

ε=

What about the time between the initial and final

times (the time between and 0=t∞=t )?

Fig. 30.12 shows increasing with time.)(ti

PHYS 153 08W 12PHYS 153 08W 12

Fig. 30.12

Fig. 30.12 shows increasing with time.)(ti

L

dt

iR

di=

−εdt

L

R

Ri

di−=

−ε

'

00 '

'

dtL

R

Ri

di ti

∫∫ −=−ε

tL

R

R

Ri

−=−

−)ln(

ε

εt

LR

e

R

Ri

)(−

=−

−

ε

ε

)1()( t

LR

eR

i−

−=∴ε

( growth in an inductor))(ti

L=τ is the time constant. When ,

Lt = I

eIti 63.0)11()( ≈−=

PHYS 153 08W 13PHYS 153 08W 13

R

L=τ is the time constant. When ,

R

Lt = I

eIti 63.0)11()( ≈−=

S2 is closed, and the battery is removed.

Why not just open S1? (S1 will have to opened

after S2 is closed to avoid shorting the battery).

When the current has reached a steady state

value 0I

Setting ,0=ε 0=−−dt

diLiR

PHYS 153 08W 14PHYS 153 08W 14

Fig. 30.13

dtL

R

i

di−= ∫∫ −= '

'

'

0

dtL

R

i

dii

I

tL

R

I

i−=

0

ln

tL

R

eIi)(

0

−=

L-C circuit

PHYS 153 08W 15PHYS 153 08W 15

Notice how this circuit behaves very differently from the R-C, or the R-L

circuits. Here there is no exponential growth, or decay. When the switch

is closed, the charge on the fully charged capacitor begins to decay, and

current flows in the circuit. The energy that was stored in the electric field

of the capacitor transfers to energy being stored in the magnetic field of

the inductor.

After , the magnetic flux is a maximum, the the magnetic field

begins to decay. The charge on the capacitor begins to build up with the

opposite polarity. This energy exchange oscillates back and forth. For an

ideal circuit (i.e. no reistance), there is no energy dissipation and the stored

energy oscillates back and forth between the capacitor and the inductor

indefinitely

Of what kind of a system, that we have studied, does this remind you?Answer: undamped SHM.

0=Φ

dt

d B

Suppose at a certain time the charge on the capacitor is discharging, so

(see text for more in-depth discussion).

PHYS 153 08W 16

Suppose at a certain time the charge on the capacitor is discharging, so

.0<dt

dqThen using Kirchhoff’s loop rule,

.0=−−dt

diL

C

q

dt

dqi =Since , then 0

12

2

=+ qLCdt

qd

Indeed, this equation has the same form as

the equation for undamped SHM.

02

2

=+ xm

k

dt

xd

Note the analogies.

xq⇔ xvi⇔

mL⇔ k⇔1

PHYS 153 08W 17

mL⇔ kC⇔1

.AQ⇔

This makes it easy to set up equations, and

to define parameters for the oscillating circuit.

(see Table 30.1)

L-R-C series circuit

What was the analysis that followed undamed SHM? Damped SHM.

Adding a resistor to the L-C circuit damps the oscillating current.

With a resistor in the circuit, Kirchhoff’s loop rule becomes,

0=−−−C

q

dt

diLiR

Replacing with and rearranging, we get idt

dq

PHYS 153 08W 18

Replacing with and rearranging, we get idt

01

2

2

=++ qLCdt

dq

L

R

dt

qd

For a solution, use the analogy to damped SHM.

02

2

=++ xm

k

dt

dx

m

b

dt

xd

Which has the solution (for )

)cos( ')2

(φω +=

−tAex

tm

b

where 2

2'

4m

b

m

k−=ω

Borrowing from this, the solution to the damped oscillatory circuit is,

kmb 42 <

This is underdamped SHM.

PHYS 153 08W 19

Borrowing from this, the solution to the damped oscillatory circuit is,

])4

1cos[(

2

2)

2(

φ+−=−

tL

R

LCAeq

tL

R

and 2

2'

4

1

L

R

LC−=ω

(for )

C

LR

42 <

(underdamped series circuit)CRL −−

When the oscillation is critically damped. C

LR 42 =

PHYS 153 08W 20

The figures show the underdamped and critically damped behaviours of

Q vs. .t

Fig.30.16

Problem 30.65

Fig. 30.25

In the circuit shown, switch

S is closed at time t = 0.

(a) Find the reading of each

meter just after S is closed.

(b) What does each meter

read long after S is closed?

PHYS 153 08W 21

Problem 30.72

In the circuit shown, neither

the battery nor the inductors

have any appreciable

resistance, the capacitors are

initially uncharged, and the

switch S has been in position

1 for a long time.

(a) What is the current in the

circuit?

PHYS 153 08W 22

circuit?

(b) The switch S is now suddenly

flipped to position 2. Find the maximum

charge that each capacitor will receive,

and how much time after the switch is

flipped that it will take them to acquire this charge.

Fig. 30.30