A coil having 300 turns and a radius of 2.0 cm carries a current of 1.0 A. If it is placed in a uniform 3.0 T magnetic field, find the torque this magnetic field exerts on the coil if the normal to the plane of the coil is oriented

The torque this magnetic field exerts on the coil is 1.131 N.m

Explanation:

Given;

Number of turns, N = 300 turns

Radius of coil, r = 2.0 cm = 0.02 m

Current in coil, I = 1.0 A

Magnetic field strength, B = 3.0 T

Torque exerted by this magnetic field is given as;

τ = N(μ x B)

Where;

μ is dipole moment

B is Magnetic field strength

N is the number of turns of the coil

μ = IA

where;

I is the current in the coil

A is the area of the coil

τ = NIABSinθ

Substitute the given values and calculate τ

τ = 300 x 1 x (π x 0.02²) x 3 x sin90

τ = 1.131 N.m

Therefore, the torque this magnetic field exerts on the coil is 1.131 N.m

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