Question

A square current loop 4.9 cm on each side carries a 520 mA current. The loop is in a 1.0 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30∘ away from the field direction.
What is the magnitude of the torque on the current loop? 2 sig fig.

Answer 1

Answer:

Torque = 0.25Nm

Explanation:

To calculate the magnitude of torque, first we must calculate the lever arm. The lever arm is perpendicular distance from the axis of rotation to the line of the action of the force.

Lever arm = r sin theta

Where r = 4.9cm= 0.049m

Theta = 30°

Lever arm = 0.049 sin30° = 0.245m

Magnitude of torque = rfsintheta

f = 1.0T

Magnitude of torque= 0.245×1.0 = 0.245Nm

To 2 significant figure=0.25Nm

Answer 2

Answer:

τ = 6.48×10-⁵ N•m

Explanation:

See attachment below