Question

A circular coil of 100 turns, radius 10 cm carries a current of 5 A. It is suspended vertically in a uniform horizontal magnetic field of 0.5 T and the field lines make an angle of 60° with the plane of the coil. The magnitude of the torque that must be applied on it to prevent it from turning is

• A. 2.93 Nm
• B. 3.43 Nm
• C. 3.93 N m
• D. 4.93 N m

Answer 1

Answer: (C)

3.93 N m

Answer 2

: Here, ,

Area of the coil, ,

$\mathrm { A } = \pi \mathrm { r } ^ { 2 } = 3.14 \times ( 0.1 ) ^ { 2 }$

$\tau = \mathrm { NIBA } \sin \theta$

$= 100 \times 5 \times 0.5 \times 3.14 \times ( 0.1 ) ^ { 2 } \times \sin 30 ^ { \circ }$

$= 3.931 \mathrm { Nm }$