Introduction to Electrodynamics: Pearson New International Edition by Griffiths David J

Author:Griffiths, David J. [Griffiths, David J.]
Language: eng
Format: epub
Publisher: Pearson Education Limited
Published: 2013-08-26T18:30:00+00:00

tention with J—that’s why Faraday and the others never discovered it in the lab-

oratory. However, it plays a crucial role in the propagation of electromagnetic

waves.

Apart from curing the defect in Ampère’s law, Maxwell’s term has a cer-

tain aesthetic appeal: Just as a changing magnetic field induces an electric field

(Faraday’s law), so21

A changing electric field induces a magnetic field.

20For the history of this subject, see A. M. Bork, Am. J. Phys. 31, 854 (1963).

21See footnote 8 for commentary on the word “induce.” The same issue arises here: Should a chang-

ing electric field be regarded as an independent source of magnetic field (along with current)? In a

proximate sense it does function as a source, but since the electric field itself was produced by charges

and currents, they alone are the “ultimate” sources of E and B. See S. E. Hill, Phys. Teach. 49, 343

(2011); for a contrary view, see C. Savage, Phys. Teach. 50, 226 (2012).

338

Electrodynamics

Of course, theoretical convenience and aesthetic consistency are only suggestive—

there might, after all, be other ways to doctor up Ampère’s law. The real confir-

mation of Maxwell’s theory came in 1888 with Hertz’s experiments on electro-

magnetic waves.

Maxwell called his extra term the displacement current:

∂E

J d ≡ 0

.

(38)

∂t

(It’s a misleading name; 0 (∂E /∂t) has nothing to do with current, except that it

adds to J in Ampère’s law.) Let’s see now how displacement current resolves the

paradox of the charging capacitor (Fig. 43). If the capacitor plates are very close

together (I didn’t draw them that way, but the calculation is simpler if you assume

this), then the electric field between them is

Q

E = 1 σ = 1

,

0

0 A

where Q is the charge on the plate and A is its area. Thus, between the plates

∂ E = 1 dQ = 1 I.

∂t

0 A dt

0 A

Now, Eq. 37 reads, in integral form,

∂E

B · dl = μ 0 I enc + μ 0 0

· da .

(39)

∂t

If we choose the flat surface, then E = 0 and I enc = I . If, on the other hand, we

use the balloon-shaped surface, then I enc = 0, but (∂E /∂t) · da = I / 0. So we get the same answer for either surface, though in the first case it comes from the

conduction current, and in the second from the displacement current.

Example 14. Imagine two concentric metal spherical shells (Fig. 44).

The inner one (radius a) carries a charge Q(t), and the outer one (radius b) an

opposite charge − Q(t). The space between them is filled with Ohmic material of

conductivity σ , so a radial current flows:

Q

σ Q

J = σ E = σ 1

ˆr; I = − ˙ Q =

J · da =

.

4 π 0 r 2

0

This configuration is spherically symmetrical, so the magnetic field has to be zero

(the only direction it could possibly point is radial, and ∇ · B = 0 ⇒

B · da =

B( 4 πr 2 ) = 0, so B = 0). What? I thought currents produce magnetic fields! Isn’t that what Biot-Savart and Ampère taught us? How can there be a

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