A square coil of side 10 cm has 20 turns a carries of current 12A.The coil is suspended vertically and normal to the plane of the coil makes an angle theta with the direction of uniform horizontal magnetic field of 0.80 T,the torque experienced by the coil equatorial 0.96 N-m,find the value of theta.
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Force on a conductor of n turns carrying current i, length l in a magnetic field (flux density)
The forces on the two horizontal sides DA and BC of the square coil are equal and in opposite directions (vertically) along the vertical axis of the coil. So they cancel. The forces F1 and F2 on the vertical sides AB and CD of the square coil are horizontal and are in opposite directions. But they are not along the same line. So they cause the torque direction vertically upwards.
Total torque = 2 r F Sin θ= 2 (l/2) F sin θ
= n i l² B sin θ
0.96 = 20 * 12 * 0.1² * 0.80 sinθ
θ = 30°
The torque can be expressed as
The forces on the two horizontal sides DA and BC of the square coil are equal and in opposite directions (vertically) along the vertical axis of the coil. So they cancel. The forces F1 and F2 on the vertical sides AB and CD of the square coil are horizontal and are in opposite directions. But they are not along the same line. So they cause the torque direction vertically upwards.
Total torque = 2 r F Sin θ= 2 (l/2) F sin θ
= n i l² B sin θ
0.96 = 20 * 12 * 0.1² * 0.80 sinθ
θ = 30°
The torque can be expressed as
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