Force Between Two Parallel Currents — Question 7

Question 7

Quasistatic separation of attractive currents requires logarithmic mechanical work

Two parallel long wires carry constant currents I1 and I2 in the same direction. One wire is fixed; an external agent slowly moves the other from separation d1 to d2, with $d_2>d_1$. Current supplies maintain both currents; end effects are neglected. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the external mechanical work per unit length.

2. Evaluate for I1=4.00 A, I2=6.00 A and d2=3d1, and state why this is not the full electrical-source energy balance.

Question 7 - Solution

Quasistatic separation of attractive currents requires logarithmic mechanical work

(a) Method and calculation

$$f_{\mathrm{ext}}(d)=\frac{\mu_0I_1I_2}{2\pi d},\quad\frac{W_{\mathrm{ext}}}\ell=\int_{d_1}^{d_2}f_{\mathrm{ext}}(d)\,dd.$$$$\boxed{\frac{W_{\mathrm{ext}}}\ell=\frac{\mu_0I_1I_2}{2\pi}\ln\frac{d_2}{d_1}}.$$

(b) Interpretation and further calculation

$$W_{\mathrm{ext}}/\ell=(4.80\times10^{-6})\ln3=\boxed{5.27334\times10^{-6}\,\mathrm{J/m}}.$$$$W_{\mathrm{magnetic,mechanical}}=-W_{\mathrm{ext}}.$$Maintaining the currents during motion also involves the current supplies; the mechanical integral alone does not determine their energy transfer.

Physical check

The logarithm is dimensionless and the mechanical work is positive when attractive wires are separated quasistatically.