W9_parallel_resonance.eps

Week 8: Faraday’s Law and Induction 285

works, but it is useful toshow howit works in an actual context from time to time to reinforce the
idea.

Butisall of that power being delivered to the inductor going? It isn’t being burned and released
as heat – that part of the tally is accounted for in the resistance! Maybe – could it be – is it possible

• that the energy is going into themagnetic field?

It is.

8.9: Magnetic Energy

Let’s imagine that the power delivered to the inductor is is somehow beingstoredin the inductor in
the magnetic field. Then:

PL=dUL

dt

=−LIdI

dt

(621)

or (multiplying bydt):

dUL = −LI dI

∫Utot

0

dUL =

∫I 0

0

−LI dI

Utot =

1

2

LI 02 (622)

This is the moral equivalent of theU=^12 CV^2 that we similarly derived for a capacitor, but this is
adynamicquantity as it depends on the currentflowingin the inductor.

Let is imagine that our inductor is an ideal solenoid withNturns, lengthℓ, and cross-sectional
areaA, one where the magnetic field inside the solenoid is constant and equal in magnitude to:

B=

μ 0 N I 0

ℓ (623)

and that vanishes at the ends of the solenoid (neglecting fringing fields). We showed above that the
self-inductance of this ideal solenoid is:

L=

μ 0 N^2 A

ℓ

(624)

Let’s do an algebra-morph of the energy stored on the inductor:

U =^1

2

LI^2

=

μ 0 N^2 A

2 ℓ

I^2

=

μ^20 N^2 Aℓ

2 ℓ^2 μ 0

I^2

=^1

2 μ 0

μ^20 N^2 I^2

ℓ^2

Aℓ

∆U =

B^2

2 μ 0

∆V

∆U

∆V =

B^2

2 μ 0 (625)

where we have used the fact thatAℓ= ∆V, thevolumeof the solenoid (the only region where our
idealized field is not zero).