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Circular coil of 30 turns, and radius 8....

Circular coil of 30 turns, and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of `60^(@)` with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil form turning.

Text Solution

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Number of turns on the circular coil, n=30.
Radius of the coil, r=8.0 cm 0.08 m
Area of the coil `pir^(2) = pi(0.08)^(2) = 0.0201m^(2)`
Current flowing in the coil, I=6.0A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface, `theta = 60^(@)`The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,
`tau = n IBA sin theta = 30xx6xx1xx 0.0201xx sin 60^(@)` = 3.133 Nm
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