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(a) A 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 counter torque that must be applied to prevent the coil from turning. (b) Whould your answer change, if the circular coil in (a) were replaced by a planer coil of some irregular shape that encloses the same area? (All other particulars are also unaltered). |
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Answer» Solution :Here, `N = 30, R = 8.0 cm = 8 xx 10^(-2) m , I = 6.0 A , theta = 60^@ and B = 1.0 T` (a) `:.` The magnitude of the counter torque = magnitude of the DEFLECTING torque `= N A I B sin theta = N cdot (PI R^2) I B sin theta` `= 30 xx 3.14 xx (8 xx 10^(-2)) xx 6.0 xx 1.0 xx sin 60^@` `= 3.14 N m`. (b) The answer would not change as area enclosed by the coil as WELL all other particulars remain unaltered and the formula `tau = N A I B sin theta` is true for PLANAR coils of any shape. |
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