A flat, 193 193 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.55 cm 2 4.55 cm2 and the angle between its magnetic dipole moment and the field is 59.1 ∘ . 59.1∘. Find the strength B B of the magnetic field that causes a torque of 1.47 × 10 − 5 N ⋅ m 1.47×10−5 N⋅m to act on the loop when a current of 2.13 mA 2.13 mA flows in it.

The speed was not exceeded so it was slow

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j