Question

A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.230 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A. (a) Find the force on each side of the loop. magnitude µN direction Correct: Your answer is correct. (b) Find the magnitude of the torque acting on the loop. 0 Correct: Your answer is correct. N · m

Answer 1

solenoid :

i_s = current in solenoid = 15 A

n = number of turns/length = 30/cm = 30 /(10^-2)m = 3000 turns/m

magnetic field by the solenoid parallel to its axis is given as

B = $\mu$ ni_s= (12.6 x 10^-7) (3000) (15) = 5.7 x 10^-2 T

a)

i = current flowing in square = 0.230 A

a = length of each side = 2 cm = 0.02 A

Force due to magnetic field inside the solenoid on each arm of square is given as

F = i Ba Sin90 = 0.23 (5.7 x 10^-2) (0.02)

F = 2.6 x 10^-4 N

Part B)

The magnetic force on each arm of the square is equal and opposite and in the same plane as the square itself. hence the forces does not make a rotating pair to rotate the square loop

hence Torque = 0