Question

A current of 19.5 mA is maintained in a single circular loop with a circumference of 2.30 m. A magnetic field of 0.710 T is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop?

Answer 1

Answer:

0.0058 Nm

Explanation:

Parameters given:

Number of turns, N = 1

Current, I = 19.5 mA = 0.0195 A

Circumference = 2.3 m

Magnetic field, B = 0.710 T

Angle, θ = 90°

Magnetic torque is given as:

τ = N * I * A * B * sinθ

Where A is area

Circumference is given as:

C = 2 * pi * r

=> 2.3 = 2 * pi * r

=> r = 2.3/(2 * 3.142)

r = 0.366 m

Area, A can now be found:

A = pi * r² = 3.142 * 0.366²

A = 0.42 m²

Therefore, torque is:

τ = 1 * 0.0195 * 0.42 * 0.71 * sin90

τ = 0.0058 Nm