A circular coil of wire 8.60 cm in diameter has 18.0 turns and carries a current

Question

A circular coil of wire 8.60 cm in diameter has 18.0 turns and carries a current of 3.20 A . The coil is in a region where the magnetic field is 0.590 T .

A)What orientation of the coil gives the maximum torque on the coil ? Please, enter the value of the angle between the field and the normal to the plane of the loop.

B)What is this maximum torque in part (A) ?

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

torque on a body is defined by the equation, T = NiAB sin theta

where N # of turns

i is current

A is area

B is magnetic field

theta

so for maximum orientation theta = 90 deg

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part B :

Tmax = 18 * 3.14* 0.086* 0.086/4 * 3.2 * 0.59

Tmax = 0.197 Nm

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apply T at 71% = 0.197 * 0.71 = 0.140

so sin theta = 0.140/0.197 =0.71

theta = 45.28 deg

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