Question

A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 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 magnetic torque on the coil?

Answer 1

A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?

Magnetic torque on the coil = 0.08796 N.m

Step-by-step explanation:

Given:

A circular coil with N turns and I current along with its diameter d which makes an angle (\theta) .

Number of turns of the coil, N = 200

External magnetic field, B = 0.7 T

Current in the circular coil, I = 4 A

Diameter of the coil, d = 2 cm = 0.02 m

Angle between the coil and B ,
`(\theta)` = 30°

Formula to be used:

Magnetic torque in a circular coil:

⇒
`\tau = NIABsin(\theta)`

⇒
`\tau = NIB((\pi d^2)/(4))sin(\theta)` ...Area of cross section = πd^2/4

⇒ Plugging the values.

⇒
`\tau = 200* 4* 0.7* (\pi (0.02)^2)/(4)* sin(30)`

⇒
`\tau = 200* 4* 0.7* (\pi (0.02)^2)/(4)* 0.5` ...sin(30)=0.5

⇒
`\tau = 0.08796\ N.m`

So,

The magnetic torque on the coil = 0.08796 N.m