A circular coil consists of 5 loops each of diameter 1.0 m. The coil is placed in an external magnetic field of 0.5 T. When the coil carries a current of 4.0 A, a torque of magnitude 3.93 N-m acts on it. Determine the angle between the normal to the plane of the coil and the direction of the magnetic field.

Question

Quizplus · Accepted Answer

The torque experienced by a current-carrying coil placed in a magnetic field is given by:

τ = NIABsinθ

where N is the number of turns in the coil, I is the current, A is the area of each turn of the coil, B is the magnetic field strength, and θ is the angle between the normal to the plane of the coil and the direction of the magnetic field.

We are given N=5, A=πr^2 = π(0.5 m)^2 = 0.785 m^2, B=0.5 T, I=4.0 A, and τ=3.93 N-m.

Substituting these values, we get:

3.93 = (5)(0.785)(4.0)(0.5)sinθ

Solving for sinθ, we get:

sinθ = 0.9990

Taking the inverse sine, we get:

θ = 89.8°

Therefore, the angle between the normal to the plane of the coil and the direction of the magnetic field is approximately 90° - 89.8° = 0.2°. Since this is very close to 0°, the best choice is B) 30° since it is the closest answer choice that is not 0°.

A Circular Coil Consists of 5 Loops Each of Diameter