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Question

(a) A circular coil of 30 turns and radius 8.0 cm carrying a currentof 6.0 A is suspended vertically in a uniform horizontal magneticfield of magnitude 1.0 T. The field lines make an angle of 60°with the normal of the coil. Calculate the magnitude of thecounter torque that must be applied to prevent the coil fromturning. (b) Would your answer change, if the circular coil in (a) were replacedby a planar coil of some irregular shape that encloses the samearea? (All other particulars are also unaltered.)

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Solution

a)

Given: The numbers of turns on the circular coil are 30, the radius of the coil is 8.0cm, the current flowing in the coil is 6.0A, the magnetic field strength is 1T and the angle between the field lines and normal with the coil surface is 60°.

The area of the coil is given as,

A=π r 2

Where, the radius of the coil is r and the area of the coil is A.

By substituting the given value in the above formula, we get

A=π ( 0.08 ) 2 =0.0201 m 2

The counter torque applied for preventing the coil from turning is given as,

T=nBIAsinθ

Where, the numbers of turns on a circular coil are n, the current flowing in the coil is I, the magnetic field strength is B and the angle between the field lines and normal with the coil surface is θ.

By substituting the given values in the above formula, we get

T=30×6×1×0.0201×sin60° =3.133Nm

Thus, the magnitude of the counter torque is 3.133Nm.

(b)

From the result obtained in part (a), it is clear that the magnitude of torque depends on the area of the coil. It does not depend on the shape of the coil.

Thus, the magnitude of the counter torque will not be changed and it remains same.


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