# (a) A circular coil of 30 turns and radius 8.0 cm 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 - Physics

Numerical

(a) A circular coil of 30 turns and radius 8.0 cm 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° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

(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.)

#### Solution

(a) Number of turns on the circular coil, n = 30

Radius of the coil, r = 8.0 cm = 0.08 m

Area of the coil = πr2 = π(0.08)2 = 0.0201 m2

Current flowing in the coil, I = 6.0 A

Magnetic field strength, B = 1 T

Angle between the field lines and normal with the coil surface,

θ = 60°

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

τ = n IBA sin θ …........(i)

= 30 × 6 × 1 × 0.0201 × sin 60°

= 3.133 N m

(b) It can be inferred from relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

Concept: Moving Coil Galvanometer
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#### APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 4 Moving Charges and Magnetism
Q 13 | Page 169
NCERT Physics Part 1 and 2 Class 12
Chapter 4 Moving Charges and Magnetism
Exercise | Q 4.13 | Page 169