Section E

\oint B \cdot d l = μ_{0} I_{enc}

. Its general vacuum version adds

μ_{0} ϵ_{0} d Φ_{E} / d t

on the right. Faraday's law is

E = \oint E \cdot d l = - d Φ_{B} / d t

for a fixed loop. A circular Amperian path around a long straight wire gives

B (r) = \frac{μ _{0} I}{2 π r},

(1)

with direction given by the right-hand rule. The force on a wire element is

d F = I d l \times B

; its magnitude is

I B d l sin χ

and its direction follows this cross product.

Choose the straight-wire current rightward and the square below it, as printed. Its positive magnetic field there is into the page. With an into-page loop normal, the magnetic flux linkage is

Λ = N \int_{s}^{s + a} \frac{μ _{0} I}{2 π r} a d r = M I, M = \frac{μ _{0} N a}{2 π} ln \frac{s + a}{s} .

(2)

Here

M

denotes mutual inductance, not a mass. Choose positive coil current to produce an out-of-page field. Then its induced current in the resistance-dominated approximation is

i (t) = \frac{M}{R} \frac{d I}{d t} = \frac{μ _{0} ω N a I _{0}}{2 π R} ln \frac{s + a}{s} cos (ω t) .

(3)

This reproduces the specified sign with the declared orientation. It neglects coil self-inductance, requiring

ω L_{coil} ≪ R

, as implicit in the requested expression.

The nearest side contains N parallel current segments and is at a uniform distance

s

, so its force magnitude is

F_{near} = \frac{μ _{0} N a}{2 π s} ∣ I (t) i (t) ∣.

(4)

The two sides perpendicular to the wire have opposite currents and equal magnetic fields at corresponding distances; their forces cancel pairwise. The far side opposes the near-side force and has the smaller field. Thus

F_{total} = \frac{μ _{0} N a}{2 π} ∣ I (t) i (t) ∣ (\frac{1}{s} - \frac{1}{s + a}) .

(5)

When

I

and

i

have the same sign under the chosen conventions, the near-side coil current opposes the straight-wire current and the total force is away from the wire; otherwise it is toward it. In particular the signed force alternates with

sin (ω t) cos (ω t)

. The coil is fixed, so these forces do no mechanical work on it. The mean Joule heating is

\overline{P} = R \overline{i^{2}} = \frac{M ^{2} ω ^{2} I _{0}^{2}}{2 R} .

(6)

E15

 0  0

a

 0  0

Solution

 0  0

A gravitational field is force per unit test mass,

g = F / m

. Newton's law of universal gravitation gives the field of a point mass M:

g = - \frac{GM}{r ^{2}} r, Φ = - \frac{GM}{r}, g = - \nablaΦ.

(1)

The potential is potential energy per unit mass, with zero at infinity for an isolated mass. Fields and potentials superpose. The potential difference fixes the work per unit mass, and a conservative gravitational field has zero circulation around a closed path.

For a continuous mass distribution, the gravitational Gauss law is

\oint g \cdot d S = - 4 π G M_{enc}

. A spherical body therefore has the same exterior field as a point mass at its centre. Inside a uniform sphere of density

ρ

g = - \frac{4 π Gρ}{3} r .

(2)

The field is continuous at its surface and vanishes at its centre. An empty spherical shell has zero interior field but a nonzero constant interior potential. Near a small region of Earth's surface one may approximate