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A closely wound solenoid of 1000 turns and area of cross-section `2 xx 10^(-4) m^(2)` carries a current of 1A. It is placed in a horizontal plane with its axis making an angle of `30^(@)` with the direction of uniform magnetic field of `0.16 T`. Calculate the torque acting on the solenoid. |
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Answer» Given : `A = 2 xx 10^(-4) m^(2),I = 1A, theta = 30^(@)` `B = 0.16 T, n = 1000` As we know, `M = nIA` `= 1000 xx 1 xx 2 xx 10^(-4)` `M = 0.2 Am^(2)` Now, `tau = MB sin theta` `= 0.2 xx 0.16 xx sin 30^(@)` `= 0.2 xx 0.16 xx 1/2` `tau = 0.016 Nm`. |
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