Chapter 30 Inductance

1 Mutual Inductance

What is the induced emf E2 in coil 2 due the change in current in coil 1, where coil1 is in close proximity to coil 2?

From Faraday’s Law, we can write the following:

E = −N2dΦB2

dt

The flux linkage in the second coil is defined to be:

N2 ΦB2 = M21 i1 (1)

where M21 is the constant of proportionality constant called the mutual induc-tance. The SI units of inductance is the henry (H), named after Joseph Henry. Ifwe look at the time-rate-of-change of Eq. 1, we find:

N2dΦB2

dt= M21

di1dt

1

Therefore, we can write the induced emf in coil 2 as:

E2 = −M21di1dt

(induced emf in the second coil) (2)

If we reverse the roles of coil 1 and coil 2, we find that M12 = M21 leading to thefollowing relationships:

E2 = −Mdi1dt

and E1 = −Mdi2dt

where the mutual inductance M is:

M =N2ΦB2

i1=

N1ΦB1

i2(definition of mutual inductance)

2 Self-Inductance and Inductors

As a current passes through the wires wrapped about a solenoid, the increasingmagnetic field creates a self-induced emf due to Lenz’s law. The induced emfis set up in such a way as to oppose the change in the current that caused theemf.

As we did for the mutual inductance, we can write the flux linkage for this cir-cuit:

NΦB = Li

where L is the constant of proportionality called the self-inductance. Again, asbefore, let’s take a derivative of the flux linkage to find the following:

NdΦB

dt= L

di

dt

From Faraday’s Law, we can now write the self-induced emf:

E = −Ldi

dt

2

The solenoid in the above circuit is what we call an inductor. The purpose of theinductor is to oppose any variations in the current through a circuit.

Vab = Va − Vb = Ldi

dt

3

2.1 Self-inductance for a toroidal solenoid

ΦB = BA =µoNiA

2πr

From our definition of the flux linkage, we can write

NΦB = Li or L =NΦB

i=

µoN2A

2πr

2.2 Self-inductance for a simple solenoid

We can write the flux linkage as

NΦB = L i ⇒ L =NΦB

i=

NBA

i=

NnµoiA

i

So, now we can write the self-inductance for a simple solenoidal inductor as:

L = NnµoA (inductance for a simple solenoid)

Note: The inductance only depends on the geometry of the inductor (i.e.,the number of turns, the material, and its physical size.

4

Ex. 7 The inductor in Fig. 30.18 has inductance 0.260 H and carries a cur-rent in the irection shown that is decreasing at a uniform rate, di/dt =−0.0180 A/s. a) Find the self-induced emf. b) Which end of the induc-tor, a or b, is at a higher potential.

3 Magnetic Field Energy

Establishing a current in an inductor builds up the magnetic field inside the in-ductor, and this requires work. The instantaneous power supplied by the externalsource to do this work is:

P = Vabi = Lidi

dt

The energy supplied by the external source in an infinitesimal amount of time dt

is:

dU = P dt = Li di

The total energy U supplied while the current increases from zero to its final valueI is:

U = L

∫ I

0i di =

1

2L I2 (energy stored in an inductor)

Ex. 12 An inductor used in a dc power supply has an inductance of 12.0 Hand resistance of 180 Ω. It carries a current of 0.300 A a) What is theenergy stored in the magnetic field? b) At what rate is thermal energydeveloped in the inductor? c) Does your answer to part (b) mean thatthe magnetic-field energy is decreasing with time? Explain.

5

3.1 Energy stored in a toroidal solenoid

Recall that we can write the inductance for a toroidal solenoid as:

L =µo N 2A

2πr

Now, we can write the total energy stored in a toroidal solenoid as:

U =1

2

µo N 2A

2πrI2

The energy per unit volume is:

u =U

V=

U

2πrA=

1

2µo

N 2I2

(2πr)2 (3)

We can rewrite this equation in terms of the magnetic field B inside the toroidalsolenoid. Recall, that the magnetic field inside a toroidal solenoid is B = µoNI/2πr.Now we can rewrite the last fraction in the previous equation as:

N 2I2

(2πr)2 =B2

µ2o

Now, we can rewrite Eq. 3 as:

u =B2

2 µo(magnetic energy density in a vacuum)

This is analogous to the electric energy density stored in a vacuum u = 12εoE

2.

If the material inside the toroid is not vacuum but a material with magnetic per-meability µ = Km µo, then the magnetic energy density becomes:

u =B2

2 µ(magnetic energy density in a material)

6

4 The R-L Circuit

The R− L circuit is shown in the figure below.

Using Kirchhoff’s rule (∑

Vi = 0), we find:

E − iR− Ldi

dt= 0

Solving this equation for the current i, we find:

i =ER

(1− e−t/(L/R)

)(current in an R-L circuit with emf)

where L/R is called the time constant for the circuit. So, once again, we havea time constant with a slightly different definition, however, the units are still inseconds.

τ =L

R

7

4.1 Current Decay in an R-L Circuit

When the inductor is discharged, the energy in the inductor is dissipated by theheat loss in the resistor. Once again, we can use Kirchhoff’s rule (

∑Vi = 0) to

obtain a differential equation. The solution to the differential equation is:

i = Io e−t/(L/R) (current decay in an R-L circuit)

5 The L-C Circuit

An L-C circuit behaves entirely different from the circuits we’ve seen so far. It’sbehavior is characterized by oscillating current and charge. Initially, we charge the

8

capacitor plate with an initial charge Q = CVm, and close the switch at t = 0.Various stages of the energy transfer are shown in the figure below.

9

In order to describe the flow of charge and current, we use Kirchhoff’s rule (∑

vi =0).

−Ldi

dt= 0

Since i = dq/dt and di/dt = d2q/dt2, we can write the above equation as:

d2q

dt2+

1

LCq = 0 (L-C circuit)

The solution to this equation is:

q = Q cos(ωot + φ)

and the angular frequency ωo is given by

ωo =

√1

LC(angular frequency in an L-C circuit)

10

6 The L-R-C Circuit

In this section we add a resistor to the previous circuit to form an L-R-C seriescircuit. The effect of the resistor is to dampen the oscillatory motion observedin the previous section. The different kinds of damping are shown in the figuresbelow.

As before, we initially charge the capacitor with an emf source E , then disconnectthe battery from the circuit, and then discharge the energy stored in the capacitorto the other two elements in the series circuit as shown below.

11

To describe the charge on the capacitor plate as a function of time, we once againinvoke Kirchhoff’s rule (

∑vi = 0) following the path abcda:

−iR − Ldi

dt− q

C= 0

Replacing i with dq/dt and rearranging, we obtain the following equation:

d2q

dt2+

R

L

dq

dt+

1

LCq = 0

The solution to this equation is rather complex, but here it is:

q = Ae−12 t/(L/R) cos

(√1

LC− R2

4L2 t + φ

)where the natural resonating frequency of the circuit (without the inductor) isωo =

√1/LC.

Homework – Chapter 30

Exercises:

Problems:

12