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.

Question

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. (A) 4 Nm (B) 6 Nm (C) 3.1 Nm (D) 2.8 Nm

Tardigrade Admin · Accepted Answer

Magnitude of torque acting on the current-carrying coil due to the magnetic field,

τ = N I A B sin θ

= 30 \times 6 \times π (0.08)^{2} \times 1 \times sin 6 0^{\circ}

= 30 \times 6 \times 3.14 \times 0.08 \times 0.08 \times \frac{3}{2}

= 3.133 N

4 Nm

6 Nm

3.1 Nm

2.8 Nm

Magnitude of torque acting on the current-carrying coil due to the magnetic field, τ=NIABsinθ =30×6×π(0.08)2×1×sin60∘ =30×6×3.14×0.08×0.08×23​​ =3.133N

Magnitude of torque acting on the current-carrying coil due to the magnetic field,

$τ = N I A B sin θ$
$= 30 \times 6 \times π (0.08)^{2} \times 1 \times sin 6 0^{\circ}$
$= 30 \times 6 \times 3.14 \times 0.08 \times 0.08 \times \frac{3}{2}$
$= 3.133 N$