(a) Find the resultant magnetic force exerted by the long wire onthe square curr

Question

(a) Find the resultant magnetic force exerted by the long wire onthe square current loop in the figure belowif I1 = 8.72 A and I2 = 3.76 A. Theedge length of the square is 19.8 cm, and the distance from the long wire tothe closest edge of the loop is 4.65 cm. magnitude N direction ---Select---towards the wireaway from the wireno direction
(b) Find the magnetic torque on the current loop. magnitude N–m direction ---Select---towards the wireaway from the wireno direction magnitude N direction ---Select---towards the wireaway from the wireno direction

Explanation / Answer

(a) we know that
               F = o L I1 I2 /2 r        
The force on the left edge is
       F1 = 4x 10-7 * 0.198 * 8.72 * 3.76 / 2 *0.0465
=   2.79 x10-5 N to the left, because currents in thesame direction attract each other. . The force on the right edge is
F2 = 4 x10-7 * 0.198 * 8.72 * 3.76 / 2 *( 0.0465 +0.198)
=   0.531 x10-5 N to the right, because oppositecurrents repel. Therefore net force F = F1 - F2 = 2.259 x10-5 N to the left

(b)The net torque on the loop is zero. This is because all theforces acting on the loop are parallel to the plane of the loop, sonone of the forces exert a torque on the loop. . The force on the right edge is
F2 = 4 x10-7 * 0.198 * 8.72 * 3.76 / 2 *( 0.0465 +0.198)
=   0.531 x10-5 N to the right, because oppositecurrents repel. Therefore net force F = F1 - F2 = 2.259 x10-5 N to the left

(b)The net torque on the loop is zero. This is because all theforces acting on the loop are parallel to the plane of the loop, sonone of the forces exert a torque on the loop.

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