A circular coil of wire 9.00cm in diameter has 17.0 turns and carries a current

Question

A circular coil of wire 9.00cm in diameter has 17.0 turns and carries a current of 2.40A . The coil is in a region where the magnetic field is 0.570T .

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.

What is this maximum torque in part (A) ?

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 = Magnetic moment X B

Magnetic moment =NIA

Torque = NIA X B =NIAB * sin(theta)

So for maxm torque,Angle between moment and magnetic field should be 90 degree

So angle = 90 degree between the field and the normal to the plane of the loop.(ans)

Torque)max = NIAB = 17*2.4* 3.14 *0.09 *0.09 *0.57 /4 = 0.147 Nm (ans)

Now for torque = 71/100 * Tmax

NIAB sin(theta) = NIAB * 0.71

sin(theta) = 0.71

theta = 45.23 degree between the field and the normal to the plane of the loop. (ans)

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