Mutual Inductance

AP Physics C

Montwood High School

R. Casao

• The magnetic flux through a circuit varies with time because of varying currents in a nearby circuit.

• This gives rise to an induced EMF through a process known as mutual induction because it depends on the interaction of two circuits.

• Consider the two closely wound coils shown in cross- section.• The current I1 in coil 1, which has N1 turns, creates magnetic field lines, some of which pass through coil 2, which has N2

turns.

• The corresponding flux through coil 2 produced by coil 1 is represented by Φ21.

• The mutual inductance M1 on 2 of coil 2 with respect to coil 1 is the ratio of N2·Φ1 on 2 and the current I1:

• The mutual inductance depends on the geometry of both circuit elements and on their orientation with respect to one another.

2 1 21 2

1

1 21 2 1

2

onon

onon

NM

I

MI

N

• As the circuit separation increases, the mutual inductance decreases since the magnetic flux linking the circuits decreases.

• If the current I1 varies with time, the EMF induced in coil 2 by coil 1 is given by:

1 22 2

1 2 11 2

2

1 2 1

22 2

on

onon

on

dEMF N

dtM I

N

M Id

NEMF N

dt

1 2 12 2

2

12 1 2

on

on

M dIEMF N

N dt

dIEMF M

dt

• If the current I2 varies with time, the induced EMF in coil 1 due to coil 2 is given by:

2 1 21 1 2 1

onon

d dIEMF N M

dt dt

• These results are similar in form to the expression for the self-induced EMF;

• The EMF induced by mutual induction in one coil is always proportional to the rate of current change in the other coil.

• If the rates at which the currents change with time are equal (that is, if dI1/dt = dI2/dt), then EMF1 = EMF2.

• Therefore, M2 on 1 = M1 on 2 = M, and

• Mutual inductance unit: Henry

dIEMF L

dt

212 1

dIdIEMF M and EMF M

dt dt

Mutual Inductance of Two Solenoids• An electric toothbrush has a base

designed to hold the toothbrush handle when not in use.

• The handle has a cylindrical hole that fits loosely over a matching cylinder on the base.

• When the handle is placed on the base, a changing current in a solenoid inside the base cylinder induces a current in a coil inside the handle.

• This induced current charges the battery in the handle.

• We can model the base as a

solenoid of length l with NB turns, carrying a source current I, and having a cross-sectional area A.

• The handle coil contains NH turns.

• Find the mutual inductance of the system.

• The base solenoid carries a source current I, the magnetic field in its interior is: o BN I

Bl

• The magnetic flux ΦBH through the handle’s coil is caused by the magnetic field of the base coil and is equal to B·A.

• Mutual inductance:

H BH

H

H o B

H o B

NM

IN B A

MI

N N I AM

I lN N A

Ml

Calculating Mutual Inductance• In one form of Tesla coil (a high-voltage generator), a

long solenoid with length l and cross-sectional area A is closely wound with N1 turns of wire. A coil with N2 turns surrounds it at its center. Find the mutual inductance.

• Mutual inductance occurs because a current in one of the coils sets up a magnetic field that causes a flux thru the other coil.

• Using

• To determine the mutual inductance, we need to know either the flux thru each turn of the outer coil due to current I1 in the

solenoid or the flux thru each turn of the solenoid due to a current I2 in the outer coil.

2 1 2 1 2 1

1 2

on onN NM

I I

• The magnetic field for a solenoid is: B = μo·(N/l)·I1

• The flux thru a cross section of the solenoid is B1·A

• A very long solenoid produces no magnetic field outside of its coil, so the flux through each turn of the outer, surrounding coil is equal to the flux thru the solenoid, no matter what the cross-sectional are of the outer coil.

2 2 2 1 2 1 1

1 1 1

2 1

B o

o

N N B A N N I AM

I I I l

N N AM

l

• Suppose l = 0.5 m, A = 0.0001 m2, N1 = 1000 turns, and N2= 10 turns:

7 2

6

10 4 10 1000 0.0001

0.5

25 10

T mx m

AMm

M x H

EMF Due to Mutual Induction• In the previous example, suppose the current I2 in the

outer, surrounding coil is given by I2 = 2 x 106 A/s·t.– At time t = 3 μs, what average magnetic flux through each

turn of the solenoid is caused by the current in the outer surrounding coil?

– Given the value of the mutual inductance from the previous problem, M = 25 x 10-6 H, determine the flux thru each turn of the solenoid caused by a given current I2 in the outer coil.

– At t = 3 μs = 3 x 10-6 s, the current in the outer coil (coil 2) is I2 = 2 x 10-6 A/s· 3 x 10-6 s = 6 A.

– The average flux through each turn of the solenoid (coil 1) is:6

7 221

1

25 10 61.5 10

1000BM I x H A

x T mN

• This is an average value for the flux; the flux can vary considerably between the center and the ends of the solenoid.

• What is the induced EMF in the solenoid?

662

1

6 61

(2 10 )25 10

25 10 2 10 50

Ad x tdI sEMF M x Hdt dt

AEMF x H x Vs