Question

What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0-A current in a 1.60-T field?

a) 2.16 N⋅m
b) 5.40 N⋅m
c) 9.60 N⋅m
d) 12.00 N⋅m

Answer 1

The maximum torque on a 150-turn square loop carrying a 50.0-A current in a 1.60-T field is calculated using the torque formula for a current loop in a magnetic field, yielding a torque of 12.00 N·m.

Step-by-step explanation:

To calculate the maximum torque exerted on a current-carrying loop in a magnetic field, we use the formula τ = NIBA sin(θ), where τ is torque, N is the number of turns, I is the current, B is the magnetic field strength, A is the area of the loop, and θ is the angle between the normal to the loop and the magnetic field. For maximum torque, the angle θ is 90°, so sin(θ) is 1. The area A of a square loop is given by A = l², where l is the length of a side of the square.

In this problem, N = 150 turns, l = 18.0 cm (which is 0.18 m), I = 50.0 A, B = 1.60 T, and sin(θ) = 1 for maximum torque. Therefore, A = (0.18 m)² = 0.0324 m². Plugging these values into the torque formula yields:

τ = (150)(50.0 A)(1.60 T)(0.0324 m²) = 12.00 N·m

So, the maximum torque on the loop is 12.00 N·m, which corresponds to option (d).