Question

A 530-turn solenoid is 13 cm long. The current in it is 37 A . A straight wire cuts through the center of the solenoid, along a 5.0-cm diameter. This wire carries a 25-A current downward (and is connected by other wires that don't concern us).

What is the force on this wire assuming the solenoid's field points due east?

Answer 1

Given

number of turns N = 530 turns

length of solenoid l = 13 cm = 0.13 m

current in solenoid I = 37 A

The magnetic field produced by solenoid is

B = muenot * N * I / l

B = 4pi * 10^-7 * 530 * 37 / 0.13

B = 0.19 T

Now a wire of length l = 5 cm is placed in this magnetic field

so the force on this wire is

F = B * I * l * sin(theta)

here B = 0.19 T

I = 25 A

l = 5 cm = 0.05 m

theta = 90 ('B' is in east and 'l' is in downward)

F = 0.19 * 25 * 0.05 * sin(90)

F = 0.2375 N