Two long parallel wires carry currents of 5.0 A and 8.0 A in the opposite direction. The wires are separated by 0.30 m. Find the magnetic force per unit length between the two wires.

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Quizplus · Accepted Answer

The magnetic force between two parallel wires carrying currents in opposite directions is repulsive. The magnitude of the force per unit length is given by:$$F = \frac{\mu_0}{4\pi} \frac{I_1 I_2}{d}$$where:* $\mu_0$ is the permeability of free space ($4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}$)* $I_1$ and $I_2$ are the currents in the wires (5.0 A and 8.0 A)* $d$ is the distance between the wires (0.30 m)Plugging in the values, we get:$$F = \frac{4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}}{4\pi} \frac{(5.0 	ext{ A})(8.0 	ext{ A})}{0.30 	ext{ m}} = 2.7 	imes 10^{-5} 	ext{ N/m}$$Therefore, the magnetic force per unit length between the two wires is $2.7 	imes 10^{-5} 	ext{ N/m}$ and it is repulsive.

Two Long Parallel Wires Carry Currents of 5