Force Between Two Parallel Currents — Question 9

Question 9

Two force measurements recover the product and sum of unknown currents

Two same-direction long currents I1 and I2 are separated by d=.0500 m. Their attraction per length is $1.60\times10^{-4}$ N/m. An independent measurement gives I1+I2=15.0 A. Both currents are positive. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the two possible current assignments.

2. If the wires have known resistances per length r1=1.00 ohm/m and r2=4.00 ohm/m, choose the assignment with lower total Joule heating per length.

Question 9 - Solution

Two force measurements recover the product and sum of unknown currents

(a) Method and calculation

$$I_1I_2=\frac{2\pi d f}{\mu_0}=40.0\,\mathrm{A^2},\quad I_1+I_2=15.0\,\mathrm A.$$$$I_{1,2}=\frac{15\pm\sqrt{225-160}}2=\boxed{\frac{15\pm\sqrt{65}}2\,\mathrm A}.$$

(b) Interpretation and further calculation

$$P_A/\ell-P_B/\ell=(r_1-r_2)(I_+^2-I_-^2)<0.$$$$\boxed{I_1=11.5311\,\mathrm A,\quad I_2=3.46887\,\mathrm A}\quad\text{minimizes }r_1I_1^2+r_2I_2^2.$$

Physical check

Magnetic force fixes the product but cannot label the wires; the resistance information resolves the energy-efficient assignment.