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 separated by a distance R carrying currents I1 and I2 is given by F/L = μ0I1I2/(2πR). In this case, the force on wire A due to the magnetic fields produced by wires B and C is F/L = μ0(24 A)(24 A)/(2π(5.0×10^-3 m)) = 6.05 N/m. Therefore, the magnitude of the force on a 4.0-m length of wire A is F = (6.05 N/m)(4.0 m) = 24.2 N = 24.2×10^-3 N = 24.2 mN. However, the force acts in a direction perpendicular to the plane of the figure and into the paper, so the correct answer is B.

The Figure Shows a Cross Section of Three Parallel Wires