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प्रश्न
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
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उत्तर
Given: Length of a side of the square coil, l = 10 cm = 0.1 m
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with the magnetic field, θ = 30°
Strength of magnetic field, B = 0.80 T
Formula: τ = nBIA sin θ
Where,
A = Area of the square coil
= l × l
= 0.1 × 0.1
= 0.01 m2
∴ τ = 20 × 0.8 × 12 × 0.01 × sin 30°
= `1.92 xx 1/2`
= 0.96 N m
Hence, the magnitude of the torque on the coil is 0.96 N m.
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