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 plane of coil makes an angle of 30° with the direction of uniform horizontal magnetic field of magnitude 0.8 T. what must be magnitude of torque experienced by the coil?
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Answer:
Length of a side of the square coil, l = 10 cm = 0.1m
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 magnetic field, θ = 30°
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
T = n BIA sinθ
Where,
A = Area of the square coil
l × l = 0.1 × 0.1 = 0.01 m2
∴ T = 20 × 0.8 × 12 × 0.01 × sin30°
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.
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Answer:
- To make 7 divisible by 8, add 0 to 7 to make it 70. Remember to place a decimal point in the quotient when this is done. As we get the remainder 0 in the third step, we cannot divide further. The answer is 0.875.
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