25 November 2002 Physics 217, Fall 2002 1

Today in Physics 217: inductance

Mutual inductanceHow mutual inductance can help with magnetic flux calculations: examples of coupled solenoids and coupled circular loops.Self inductance and back EMFUnitsCalculation of inductance

2d

1d

r

1B

I

25 November 2002 Physics 217, Fall 2002 2

Mutual inductance

Consider two coupled wire loops, arranged such that a current through one of them produces a magnetic field that has flux through the other one. For instance,

But

the proportionality factor is a constant that depends only upon the geometry of the loops.

2d

1d

r

1B

I

( )2 1 2 1 2

1 2 .

B d d

d

Φ = ⋅ = × ⋅

= ⋅

∫ ∫∫

B a A a

A

—

1 11 , soI d

c= ∫A

r1 2 2

2 1 ;BI d d Ic

⋅Φ = ∝∫ r

25 November 2002 Physics 217, Fall 2002 3

Mutual inductance (continued)

Now suppose we exchange the roles of the loops, and run a current through loop #2. We’d get the same result with just the 1s and 2s switched:

The proportionality factor is the same in the two cases. Define it as follows:

is called the mutual inductance of the loops.

2 2 21 2 .B

I d d Ic

⋅Φ = ∝∫ r

1 2 2 1

02 2 2 22

, , where1 in MKS

4

B BcMI cMId d d dM

cµπ

Φ = Φ =

⋅ ⋅ = = ∫ ∫r r

25 November 2002 Physics 217, Fall 2002 4

How mutual inductance can help with magnetic flux calculations

This simple exercise has a profound conclusion: no matter what the shape of two loops, the flux through loop 2, resulting from current I in loop 1, is exactly the same as the flux through loop 1 due to current I in loop 2.

This occasionally is useful in calculations of flux and related quantities: switching the currents and fluxes for the calculation, and then switching back, can save a lot of trouble.Consider Example 7.10: a short solenoid (length , radius a, n1 turns per unit length) and a very long one (b, n2) are coaxial. Current I flows in the short one. What is the flux through the long one?

25 November 2002 Physics 217, Fall 2002 5

Coupled solenoids

This would be difficult if calculated the way it’s stated, but easy if one instead runs the same current through the outer coil and calculates the flux through the inner:

a

b

22 1

4 .B n I n acπ

πΦ = ×

25 November 2002 Physics 217, Fall 2002 6

Coupled circular loops

A small loop of wire (radius a) lies a distance z above the center of a large loop (radius b). The planes of the two loops are parallel, and perpendicular to the common axis.(a) Suppose current I flows in the big loop. Find the flux in the little loop. (b) Suppose current I flows in the little loop. Find the flux through the big loop. (c) What is the mutual inductance?

a

b

z2 2

R

z b

=

+

25 November 2002 Physics 217, Fall 2002 7

Coupled circular loops (continued)

(a) We saw long time ago (Example 5.6, lecturenotes for 4 November 2002) that

so if the loop is small enough to consider B to be constant over itsextent,

a

b

z2 2

R

z b

=

+

( )2

3 22 2

2 ˆ ,bI bc z b

π=

+B z

( )2 2 2

3 22 2

2 .Ba b a b a bb ad B A I

c z b

πΦ = ⋅ ≅ =

+∫B a

25 November 2002 Physics 217, Fall 2002 8

Coupled circular loops (continued)

(b) Supposing that the small loop is small enough to treat as a dipole,

Then

We can compute the flux over any surface that has the loop with radius b as its boundary (Problem 7.9, on this week’s homework); looks easiest if we take it over the sphere with radius R centered in the center of the small loop:

a

b

z2 2

R

z b

=

+

3 3

2

2 cos sin ˆˆ ,

1where .

a

a

m mr r

m a Ic

θ θ

π

= +

=

B r θ

.Bb a bdΦ = ⋅∫B a

25 November 2002 Physics 217, Fall 2002 9

Coupled circular loops (continued)

a

b

z2 2

R

z b

=

+

( )

( )

( ) ( )

0 0

0

23

2

0 0

200

2 20 0

2 22 2 2 2

2 2 3

2 2 2 2

3 2 3 22 2 2 2

2 cos sin

2 2 cos sin sin

cos cos 2

cos sin

2 2 .

Bb a b

a

md R d dR

m md uduR R

m m muR R R

m mR Rm m z b m z a z bR R R R R

mb a b Icz b z b

θ θ

θ

θθ θ φ

π πθ θ θ

π π πθ

π πθ θ

π π π

π π

Φ = ⋅ =

= =

= − = −

= − −

= − − = + − +

= =+ +

∫ ∫

∫ ∫

B a

25 November 2002 Physics 217, Fall 2002 10

Coupled circular loops (continued)

(c) Thus

Note that it’s much easier to getthe fluxes -- and anything related tothem, such as induced currentsand fields -- by means of part a than part b. Replace for the MKS answer.

a

b

z2 2

R

z b

=

+

( )2 2 2

2 3 22 2

2 .Ba Bb

b a

b aMcI cI c z b

πΦ Φ= = =

+

201 with 4c µ π

25 November 2002 Physics 217, Fall 2002 11

Transformers

Back to our original loops. Suppose that the current in loop 1 varies with time. Then there is anEMF induced in loop 2:

Thus

a time-varying current in one loop ends up inducing a time-varying current in the other, with relative size M.

2d

1d

r

1B

I

2 12

1 .Bd dIMc dt dt

Φ= − = −E

2 12 ;dIMI

R R dt= = −E

25 November 2002 Physics 217, Fall 2002 12

Self inductance and back EMF

What about one loop by itself? When onetries to change I, one changes the fluxof the loop’s own B through itself, andthus produces an opposing EMF.

d

d ′

B

r

2

0

, where

1

in MKS .4

B d d

I d d cLIc

d dLc

d dµπ

Φ = ⋅ = ⋅

′ ⋅= ≡

′ ⋅=

′ ⋅ =

∫ ∫∫∫

∫∫

∫∫

B a A

r

r

r

Self inductance

25 November 2002 Physics 217, Fall 2002 13

Self inductance and back EMF

This opposing EMF,

is a form of inertia in electrical circuits. If one tries to change a current, a back EMF in induced. We will see soon that in the “equations of motion” for AC circuits, L plays the role that m plays in classical-mechanical equations.Work per unit charge done against a back EMF is just -E:

1 ,Bd dILc dt dt

Φ= − = −E Back EMF

21 .2

dW dIP I LIdt dt

W dW L IdI LI

= = − =

= = =∫ ∫

E

2

Compare:1 for capacitors.2

W CV=

25 November 2002 Physics 217, Fall 2002 14

Units

Common values in the lab:

[ ] [ ] 2 -1 sec cm in cgs units;

volt sec henry in MKS units.amp

QrL

dI Q tdt t

= = =

= ≡

E

11 2 -111 henry 10 sec cm9

−⇔ ×

15 18 2 -1

1 mH - 1 H

10 10 sec cm .

L µ− −

≈

≈ −

25 November 2002 Physics 217, Fall 2002 15

Calculation of inductance

We have the two formulas,

which turn out to be impractical in most cases for computing inductances. Far easier is the procedure resembling the calculation of capacitance: run a current, calculate the Bproduced, then the flux, and then use

as we did in the example of the coupled circular loops.

2 22 21 1, ,d d d dM L

c c′⋅ ⋅

= =∫ ∫r r

1 2 or ,B BcMI cLIΦ = Φ =

25 November 2002 Physics 217, Fall 2002 16

Calculation of inductance (continued)

For example: what’s the inductance per unit length of a very long solenoid with length , radius a, and turns per unit length n? First run a current I through it, then the magnetic field inside is

2 2 22

2 2 22 2

02

4

4

in MKS .

B

B nIc

a nBA nI n a I cLIc c

a nL a nc

π

π ππ

πµ π

=

4⇒ Φ = = ⋅ = =

4 ⇒ = =