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Home » (a) A circular coil having radius as 8.0 cm, number of turn as 30 and carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of  with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter-torque that must be applied. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
(a) A circular coil having radius as 8.0 cm, number of turn as 30 and carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60^{o} with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter-torque that must be applied. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
(a) A circular coil having radius as 8.0 cm, number of turn as 30 and carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60^{o} with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter-torque that must be applied. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

(a) Number of turns on the circular coil (n) is given as 30

Radius of the coil (r) is given as 0.08 m

Area of the coil:

\pi {r^2}= \pi {(0.08)^2}= 0.0201{m^2}

Current flowing in the coil (I) = 6.0 A

Magnetic field strength, B = 1 T

The angle formed by the field lines and the normal with respect to the coil surface., \theta = {60^ \circ }

The coil is subjected to a torque in the magnetic field. As a result, it turns. The counter torque used to keep the coil from rotating is provided by the relationship,

\tau = nIBA\sin \theta

= 30 \times 6 \times 1 \times 0.0201 \times \sin {60^ \circ }

= 3.133 N m

(b) The relationship \tau = nIBA\sin \theta indicates that the amount of the applied torque is not affected by the form of the coil. It is determined by the coil’s area. As a result, the answer remains unchanged if the circular coil in the preceding instance is replaced by a planar coil of any irregular shape enclosing the same area.

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