Force Between Two Parallel Currents — Question 6

Question 6

A rectangular loop near a straight wire experiences an unbalanced pair of parallel forces

A long wire lies on the y axis carrying I upward. A rectangular loop in the xy plane spans x=a to x=a+w and has height ell. Its current i runs upward on the near side x=a and downward on the far side. Ignore the loop self-field. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the net force on the loop due to the long wire.

2. For w much smaller than a, derive the leading force and explain the cancellation of the top and bottom forces.

Question 6 - Solution

A rectangular loop near a straight wire experiences an unbalanced pair of parallel forces

(a) Method and calculation

$$\vec f_{\mathrm{near}}=-\frac{\mu_0Ii\ell}{2\pi a}\hat x,\quad\vec f_{\mathrm{far}}=+\frac{\mu_0Ii\ell}{2\pi(a+w)}\hat x.$$$$\boxed{\vec F=-\frac{\mu_0Ii\ell}{2\pi}\left(\frac1a-\frac1{a+w}\right)\hat x}.$$

(b) Interpretation and further calculation

$$\frac1a-\frac1{a+w}=\frac{w}{a(a+w)}\simeq\frac w{a^2}\Rightarrow\vec F\simeq-\frac{\mu_0Ii\ell w}{2\pi a^2}\hat x.$$$$\vec F_{\mathrm{top}}+\vec F_{\mathrm{bottom}}=0\quad\text{because their equal field profiles meet opposite }d\vec\ell.$$

Physical check

The near attractive side wins over the far repulsive side because the long-wire field decreases with distance.