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?
A) 15 mN
B) 77 mN
C) 59 mN
D) 12 mN
E) 32 mN

Question

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? 
A) 15 mN 
B) 77 mN 
C) 59 mN 
D) 12 mN 
E) 32 mN

Quizplus · Accepted Answer

The force per unit length between two parallel wires carrying currents is given by the equation $F/L=\mu_{0}I_{1}I_{2}/2\pi R$, where $F$ is the force, $L$ is the length of one of the wires, $\mu_{0}$ is the permeability constant, $I_{1}$ and $I_{2}$ are the currents, and $R$ is the distance between the wires. The force per unit length on wire A due to wire B is $F/L=\mu_{0}(24	ext{ A})(24	ext{ A})/2\pi(5.0	imes10^{-3}	ext{ m})$, and the direction is to the left. The force per unit length on wire A due to wire C is $F/L=\mu_{0}(24	ext{ A})(24	ext{ A})/2\pi(5.0	imes10^{-3}	ext{ m})$, and the direction is to the right. Thus, the net force per unit length on wire A is zero. However, wire A is not in between wire B and wire C. One 4.0-m length of wire A experiences the same force as two unit lengths of wire A. Therefore, the magnitude of the force on the 4.0-m length of wire A is $2(24	ext{ A})(4.0	ext{ m})(2.0	imes10^{-7}	ext{ N/A}^{2})=77	ext{ mN}$.

The Figure Shows a Cross Section of Three Parallel Wires