Chapter 30 – Inductance

I.Mutual Inductance

A.Look at two sets of loops with a common axis - loop 1 has N1 turns of wire and a current I1, and loop 2 has N2 turns of wire and a current I2.

Faraday's Law says that if the current changes in one of the loops, then an induced emf will be produced in the other loop.

B.According to Faraday's Law, an emf (2) will be induced in loop 2 when the flux changes in loop 2. That is,

2 = - ,

where N2B2 is the flux through loop 2 produced by loop 1. The flux N2B2 is then dependent on the magnetic field (B1) produced by loop 1, which in turn is dependent on the current (I1) in loop 1, that is,

N2B2 B1 I1 ,

and also then

.

So, 2 , or when a constant of proportionality is introduced, 2 = - M21 . The constant M21 is called the “mutual inductance” between loops 1 and 2 due to a change in current in loop 1. In a like fashion, there is an induced emf in loop 1 that is caused by the change in current in loop 2. This emf is 1 = - M12 , where M12 is called the “mutual inductance” between loops 1 and 2 due to a change in current in loop 2.

Note that the units of mutual inductance are: [M] = volt sec/amp = henry (H).

  1. As an example find the mutual inductance between two concentric, ideal solenoids: Given two concentric, ideal solenoids. The inner solenoid has N1 turns, current I1 , length L, and radius R1 . The outer solenoid has N2 turns, current I2 , length L, and radius R2 .

1)Find the emf induced in loop 2 due to a current change in loop 1, and from this expression find the mutual inductance, M21 .

2)Find the emf induced in loop 1 due to a current change in loop 2, and from this expression find the mutual inductance, M12 .

3)Note the relationship between M21 and M12 and the fact that they do not depend on current or voltage, only the geometry of the system.

II.Self inductance

A.If current changes in a loop with N turns of wire, then the magnetic field inside the loops must be changing. If the magnetic field is changing, then according to Faraday's law there must be an induced emf in the loop. That is,

 = - (N) .

But N B I , and (N) , and therefore  . Introducing a proportionality constant, we get

 = - L ,

where L is called the "self inductance" or simply the “inductance” of the loop. The units of self inductance are the same as the mutual inductance - henries.

B.This emf  is sometimes called the "back emf" because it produces an emf that opposes the change in the current. That is, if the current in the loops is increasing, then the emf produces an induced current that opposes this increase (opposite to direction of current flow). If the current in the loops is decreasing, then the emf produces an induced current that opposes this decrease (same as direction of current flow).

C.As an example, find the self inductance of an ideal solenoid: What is the induced emf in the solenoid when the current is changing at the rate of dI/dt? The solenoid has a radius a, length b, and number of turns per length n.

D.The Inductor

1.A circuit element that is specifically designed to have inductance.

2.Circuit symbol:

3.What is the voltage across an inductor?

4.What does an inductor do?

5.What happens when current is first delivered to the inductor?

6.What happens when current flowing in the inductor is cut?

7.How much power is delivered to or supplied by the inductor?

8.How much energy is stored in the inductor?

9.What is the energy density in the magnetic field inside the inductor?

III.The Inductor in Circuits

A.The RL Series Circuit. A resistor and inductor are connected in series.

1.Switch up at time t = 0.

2.Switch down at time t = 0. The current = Io at time t = 0.

B.A more complex circuit.

1.Immediately after the switch is closed, find the current flowing through the resistors, R and R’.

2.After the switch has been closed for a long time, find the current through the resistors.

3.Assume the switch has been closed for a long time. Now open the switch. Immediately after the switch is opened, find the current through the resistors.

C.LC Series Circuit. Close switch at time t = 0. Let the capacitor have an initial charge Qo and an initial zero current.

  1. Make a qualitative analysis of charge and current.

2.Find the charge on the capacitor as a function of time.

3.Find the current in the loop as a function of time.

4.Find the energy in the capacitor and the inductor as a function of time.

D.RLC Series Circuit. At time t = 0, Q = Qo and I = 0. Close the switch at time t = 0. Find the charge on the capacitor as a function of time.

Case I: R = 0, undamped

Q = Qo cos t

Case II:R small, , underdamped

where

Case III: , critically damped

Case IV: , overdamped

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