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

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 N m (B) 3.43 N m (C) 3.93 N m (D) 4.93 N m

Tardigrade Admin · Accepted Answer

Here, N = 100, r = 10 cm = 0.10 m, I = 5 A, B = 0.5 T, θ = 90° - 60° = 30° Area of the coil, A = π r2 = 3.14 × (0.1)2 τ = NIBAsinθ = 100 × 5 × 0.5 × 3.14 × (0.1)2 × sin 30° = 3.931 N m

$2.93 N m$

$3.43 N m$

$3.93 N m$

$4.93 N m$

Here,
$N = 100$ ,
$r = 10 c m = 0.10 m$ ,
$I = 5 A$ ,
$B = 0.5 T$ ,
$θ = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$
Area of the coil,
$A = π r^{2} = 3.14 \times (0.1)^{2}$
$τ = N I B A s in θ$
$= 100 \times 5 \times 0.5 \times 3.14 \times (0.1)^{2} \times s in 3 0^{\circ}$
$= 3.931 N m$

Q. A circular coil of 100 turns, radius 10cm carries a current of 5A. It is suspended vertically in a uniform horizontal magnetic field of 0.5T 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

2.93Nm

3.43Nm

3.93Nm

4.93Nm

Here, N=100, r=10cm=0.10m, I=5A, B=0.5T, θ=90∘−60∘=30∘ Area of the coil, A=πr2=3.14×(0.1)2 τ=NIBAsinθ =100×5×0.5×3.14×(0.1)2×sin30∘ =3.931Nm

$2.93 N m$

$3.43 N m$

$3.93 N m$

$4.93 N m$

Here,
$N = 100$ ,
$r = 10 c m = 0.10 m$ ,
$I = 5 A$ ,
$B = 0.5 T$ ,
$θ = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$
Area of the coil,
$A = π r^{2} = 3.14 \times (0.1)^{2}$
$τ = N I B A s in θ$
$= 100 \times 5 \times 0.5 \times 3.14 \times (0.1)^{2} \times s in 3 0^{\circ}$
$= 3.931 N m$