The figure shows a cross section of three parallel wires each carrying a current of 24 A. The currents in wires B and C are out of the paper, while that in wire A is into the paper. If the distance R = 5.0 mm, what is the magnitude of the force on a 4.0-m length of wire A?

Question

Quizplus · Accepted Answer

The force per unit length on wire A due to wires B and C can be calculated using the equation:
$F_{AB} = \frac{\mu_0 I_A I_B}{2\pi R}$

$F_{AC} = \frac{\mu_0 I_A I_C}{2\pi R}$

where $\mu_0$ is the magnetic constant.

The forces due to wires B and C are equal in magnitude but opposite in direction, so they cancel each other out. Therefore, the net force on wire A is:

$F_A = F_{AB} - F_{AC} = \frac{\mu_0 I_A}{2\pi R} (I_B - I_C)$

Substituting the given values, we get:

$F_A = \frac{4\pi 	imes 10^{-7} 	imes 24}{2\pi 	imes 5 	imes 10^{-3}} (24 - (-24)) = 77 	ext{ mN}$

Therefore, the answer is (B).

The Figure Shows a Cross Section of Three Parallel Wires