Force Between Two Parallel Currents — Question 5

Question 5

Two suspended repelling wires determine the current through a geometric force balance

Two equal long wires each have mass per length lambda and carry equal opposite currents I. They hang from a common horizontal support line on strings of length L, one on each side, at angle theta to vertical. The wires remain parallel and level, so their separation is $d=2L\sin\theta$. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Derive I in terms of lambda, g, L and theta.

2. Evaluate for lambda=.00100 kg/m, L=.100 m, theta=10.0 degrees and g=9.81 m/s squared.

Question 5 - Solution

Two suspended repelling wires determine the current through a geometric force balance

(a) Method and calculation

$$T\cos\theta=\lambda g,\quad T\sin\theta=\frac{\mu_0I^2}{2\pi d}\Rightarrow\lambda g\tan\theta=\frac{\mu_0I^2}{2\pi d}.$$$$\boxed{I=\sqrt{\frac{4\pi\lambda gL\sin\theta\tan\theta}{\mu_0}}}.$$

(b) Interpretation and further calculation

$$d=2(.100)\sin10^\circ=.0347296\,\mathrm m.$$$$I=\sqrt{\frac{2\pi(.0347296)(.00100)(9.81)\tan10^\circ}{\mu_0}}=\boxed{17.3312\,\mathrm A}.$$

Physical check

T denotes support force per unit wire length. The horizontal repulsion is balanced by the horizontal components of the string forces.