Question

12. A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?

Answer 1

The magnitude of the magnetic torque on the given coil is 0.198 N·m.

Step-by-step explanation:

The magnetic torque on a flat circular coil can be calculated using the formula:

τ = NABsinθ

Where:

τ is the torque

N is the number of turns in the coil (200 turns)

A is the area of the coil (πr^2, where r is the radius of the coil)

B is the magnetic field strength

θ is the angle between the plane of the coil and the magnetic field

To find the magnitude of the magnetic torque on the given coil:

Radius of the coil (diameter/2): 6.0 cm / 2 = 3.0 cm = 0.03 m

Area of the coil: π(0.03 m)^2 = 0.00283 m²

Magnetic torque: τ = (200 turns)(0.00283 m²)(7.0 A)(0.5 T)sin(30°)

τ = 0.198 N·m

Answer 2

The magnitude of the magnetic torque on the coil is 1.98 A.m²

Step-by-step explanation:

Magnitude of magnetic torque in a flat circular coil is given as;

τ = NIASinθ

where;

N is the number of turns of the coil

I is the current in the coil

A is the area of the coil

θ is the angle of inclination of the coil and magnetic field

Given'

Number of turns, N = 200

Current, I = 7.0 A

Angle of inclination, θ = 30°

Diameter, d = 6 cm = 0.06 m

A = πd²/4 = π(0.06)²/4 = 0.002828 m²

τ = NIASinθ

τ = 200 x 7 x 0.002828 x Sin30

τ = 1.98 A.m²

Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²