A current carrying coil of 200 turns has an area of cross-section 1 x 10^–4 m². When suspended freely through its centre, it can turn in a horizontal plane. What is the magnetic moment of the coil for a current of 5 A? Also calculate the net force and torque on coil if a uniform horizontal field of 10 x 10^–2 T is set up at an angle of 30° with axis of coil when it is carrying the same current.

Solution

Here,
Number of turns, N = 200
Area of cross-section. A = 1 x 10^–4 m²Current passing through the coil, I = 5 A
M =?
Magnetic momemnt of the coil, M = N IA
= 200 x 5 x 10^–1
= 10^–1 JT^–1In a uniform horizontal field B = 10 x 10^–2 T, the ends of coil experience equal and opposite forces. Therefore, net force on the coil = 0.
$θ = 30^{o}$

Therefore, torque acting on the coil is,

$Torque, τ = MB \sin θ = 10^{- 1} \times 10 \times 10^{- 2} \sin 30 ° = 10^{- 2} \times \frac{1}{2} = 0.5 \times 10^{- 2} Nm$

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