Question

A square current loop 4.5 cm on each side carries a 510 mA current. The loop is in a 0.60 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30∘ away from the field direction.

Answer 1

You have not mentioned, what the question is asking for. I think, you have forgot to type the last part of the question. In my opinion, you are asking for the torque on the current loop so I am calculating the torque.

The expression for the torque is given as -

τ = µ x B
τ = IA x B
||τ|| = IABsinθ

Now, given that -

I = 510 mA = 0.510 A

A = (0.045)^2 m^2

B = 0.60 T

= 30^o

Put the values-

||τ|| = (0.510 A)(0.045 m)^2(0.60 T)sin30°
||τ|| = 3.01 x 10^-4 N-m (Answer).