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    				| 1. | A square coil of side `10cm` consists of 20 turns and carries a current of 12A. The coil is suspended vertically and normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field of magnitude `0*80T`. What is the magnitude of torque experienced by the coil? | 
| Answer» Length of a side of the square coil, `l = 10 cm = 0.1m` Current flowing in the coil, `I = 12A` Number of turns on the coil, `n = 20` Angle made by the plane of the coil with magnetic field, `theta = 30^(@)` Strength of magnetic field, `B = 0.80T` Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation, `T = n BIA sin theta` Where, `A =` Area of the square coil `rArr l xxl = 0.1 xx 0.1 = 0.01 m^(2)` `:. T = 20 xx 0.8 xx 12 xx 0.01 xx sin 30^(@)` `= 0.96N` Hence, the magnitude of the torque experienced by the coil is `0.96Nm`. | |