Question

A square current loop 5.20 cm on each side carries a 520 mA current. The loop is in a 1.30 T uniform magnetic field. The axis of the loop, perpendicular to the What is the magnitude of the torque on the current loop?

Answer 1

The magnitude of the torque on a current loop can be calculated using the formula: τ = NIAB sinθ, where τ is the torque, N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the field and the loop. In this case, a square loop with 5.20 cm sides, carrying a current of 520 mA, in a 1.30 T magnetic field, the magnitude of the torque is 359.12 x 10^-3 Nm.

Step-by-step explanation:

The magnitude of the torque on a current loop can be calculated using the formula:

τ = NIAB sinθ

Where τ is the torque, N is the number of turns in the loop, I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the magnetic field and the plane of the loop.

In this case, we have a square loop with 5.20 cm sides, carrying a current of 520 mA. The area of the loop is (5.20 cm)^2 = 27.04 cm^2. The magnetic field strength is 1.30 T, and the angle between the field and the loop is 90 degrees. Plugging these values into the formula, we get:

τ = (520 mA)(27.04 cm^2)(1.30 T) sin(90 degrees) = 359.12 x 10^-3 Nm

Therefore, the magnitude of the torque on the current loop is 359.12 x 10^-3 Nm.