Classical Electrodynamics · Problem 5.25

Maxwell Equations, Macroscopic Electromagnetism, Conservation Laws — John David Jackson

Physics — E&M · loop-wire-interaction-energy-force

Problem

A flat right rectangular loop carrying a constant current

I_1

is placed near a long straight wire carrying a current

I_2

. The loop is oriented so that its center is a perpendicular distance

d

from the wire; the sides of length

a

are parallel to the wire and the sides of length

b

make an angle

\alpha

with the plane containing the wire and the loop's center. The direction of the current

I_1

is the same as that of

I_2

in the side of the rectangle nearest the wire. (a) Show that the interaction magnetic energy

W_{12} = \int \mathbf{J}_1 \cdot \mathbf{A}_2\, d^3x = I_1 F_2

(where

F_2

is the magnetic flux from

I_2

linking the rectangular circuit carrying

I_1

), is

W_{12} = \frac{\mu_0 I_1 I_2 a}{4\pi}\ln\left[\frac{4d^2 + b^2 + 4d\,b\cos\alpha}{4d^2 + b^2 - 4d\,b\cos\alpha}\right]

(b) Calculate the force between the loop and the wire for fixed currents. (c) Repeat the calculation for a circular loop of radius

a

, whose plane is parallel to the wire and makes an angle

\alpha

with respect to the plane containing the center of the loop and the wire. Show that the interaction energy is

W_{12} = \mu_0 I_1 I_2\,d\cdot\text{Re}\{e^{i\alpha} - \sqrt{e^{2i\alpha} - a^2/d^2}\}

Find the force. (d) For both loops, show that when

d \gg a,b

the interaction energy reduces to

W_{12} \approx \mathbf{m} \cdot \mathbf{B}

, where

\mathbf{m}

is the magnetic moment of the loop. Explain the sign.