Problem 20.36 A circular coil of wire 8.80cm in diameter has 16.0 . Turns and ca

Question

Problem 20.36 A circular coil of wire 8.80cm in diameter has 16.0 . Turns and carries a current of 2.70 A. The coil is in a region where the magnetic field is 0.650 T . Please, enter the value of the angle between the field and the normal to the plane of the loop. Part B What is this maximum torque in part (A)? Part C For what orientation of the coil is the magnitude of the torque 71.0 % of the maximum found in part (B)? Please, enter the value of the angle between the field and the normal to the plane of the loop.

Explanation / Answer

A) phi = 90 degrees

Bacuse, Torque on a loop = N*A*I*B*sin(theta) (here N is no of loops, A is area of loop, I is current through loop, B is magnetic field and theta is the angle between B and normal to the plane)

B) T_max = N*A*I*B*sin(90)

= 16*pi*0.088^2*2.7*0.65

= 0.683 N.m

C) T = 71% of Tmax

N*I*A*B*sin(theta) = 0.71*N*I*A*B*sin(90)

sin(theta) = 0.71

theta = sin^-1(0.71)

= 45.2 degrees

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