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A 0.8 m length of wire is formed into a single turn, square loop in which there is a current of 12 A. The loop is placed in magnetic field of 0.12 T, which is perpendicular to a normal to the loop plane. What is the maximum torque that the loop can experience?

Respuesta :

Answer:

0.073 N-m

Explanation:

i = 12 A, l = 0.8 m, B = 0.12 T

The circumference of the loop is 0.8 m.

Let r be the radius of the loop.

2 x 3.14 x r = 0.8

r = 0.127 m

Maximum Torque = i x A x B

Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m

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