Question

A 48.0-turn circular coil of radius 5.50 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.480 T. If the coil carries a current of 23.3 mA, find the magnitude of the maximum possible torque exerted on the coil.

Answer 1

Given Information:

Magnetic field = B =  0.480 T

Current = I = 23.3 mA = 0.0233 A

Number of turns = N = 48 turns

Radius = r = 5.50 cm = 0.055 m

Required Information:

Maximum possible torque = τ = ?

Answer:

Maximum possible torque = 0.0051 N.m

Explanation:

We know that toque τ is given by

τ = NIABsin(θ)

Where N is the number of turns of the circular coil, I is the current flowing through the circular coil, A is the area of circular coil, B is the magnetic field induced in the circular coil.

The area of the circular coil is

A = πr²

A = π(0.055)²

A = 0.009503 m²

The maximum torque is possible when θ = 90°

τ = 48*0.0233*0.009503*0.480*sin(90°)

τ = 0.0051 N.m

Answer 2

Answer:

Explanation:

Given that,

Number of turn is 48

N=48

Radius is 4.8cm

r=0.048m

Magnetic Field

B=0.48T

Current in coil

i=23.3mA

i=0.233A

Maximum Torque?

Maximum torque occur at angle 90°

Torque is given as

τ = N•I•A•B•sinθ

Where N is number of turn =48

I is current in coil =0.233A

A is area of circular coil form

Area of a circle is given as

A=πr²

A=π×0.048²

A=0.007238m²

B is magnetic field =0.48T

Maximum torque occurs at 90°

τ = N•I•A•B•sinθ

τ=48×0.233×0.007238×0.48×Sin90

τ = 0.0389Nm

This torque is large enough to exert the coil