Force Between Two Parallel Currents — Question 8

Question 8

A fixed total current has one split that maximizes the attraction

Two parallel wires a distance d apart share a prescribed positive total current $I_T=I_1+I_2$, with both currents in the same direction. Neglect heating constraints and hold the geometry fixed. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the current split that maximizes the attractive force per unit length.

2. If $I_1=I_T/2+\delta$ and $I_2=I_T/2-\delta$, find the fractional force reduction and the allowed imbalance for no more than 4.00 percent reduction.

Question 8 - Solution

A fixed total current has one split that maximizes the attraction

(a) Method and calculation

$$f=\frac{\mu_0}{2\pi d}I_1(I_T-I_1),\quad\frac{df}{dI_1}=\frac{\mu_0}{2\pi d}(I_T-2I_1).$$$$\boxed{I_1=I_2=I_T/2,\quad f_{\max}=\frac{\mu_0I_T^2}{8\pi d}};\quad f\prime\prime<0.$$

(b) Interpretation and further calculation

$$f=f_{\max}\left(1-\frac{4\delta^2}{I_T^2}\right),\quad\frac{f_{\max}-f}{f_{\max}}=\frac{4\delta^2}{I_T^2}.$$$$4\delta^2/I_T^2\le.0400\Rightarrow\boxed{|\delta|\le.100I_T}.$$

Physical check

The maximum is broad: a small current imbalance changes the force only to second order.