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 on a length of wire in a magnetic field is given by:

F = ILB sinθ

where I is the current, L is the length of wire, B is the magnetic field, and θ is the angle between the wire and the magnetic field.

In this case, wire A is perpendicular to the magnetic field generated by wires B and C. The magnetic field at a distance R from a wire carrying a current I is given by:

B = μ0I/(2πR)

where μ0 is the permeability of free space.

Using this equation, we can find the magnetic field that wire A is experiencing due to wires B and C:

B = μ0(24 A)/(2π)(5.0×10^-3 m) = 3.8×10^-4 T

Therefore, the force on a 4.0-m length of wire A is:

F = (24 A)(4.0 m)(3.8×10^-4 T) = 0.036 N = 36 mN

However, we need to remember that wire A is carrying current in the opposite direction to wires B and C. This means that the force it experiences is in the opposite direction to the force on wires B and C. Therefore, the magnitude of the force on wire A is:

|F| = 36 mN

which is closest to choice B (77 mN).

The Figure Shows a Cross Section of Three Parallel Wires