The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and carries a current 10 A suspended vertically and the normal to the plane of coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.9 T is

Question

The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and carries a current 10 A suspended vertically and the normal to the plane of coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.9 T is (A) 1.6 N m (B) 1.2 N m (C) 1.4 N m (D) 1.8 N m

Tardigrade Admin · Accepted Answer

τ =NIA B sin θ Here, N=25, I=10 A, B=0.9 T, θ =30° A=a2=12 × 10-2 × 12× 10-2 144 × 10-4 m2 ∴ τ=25 × 10 × = 144× 10-4× 0.9× sin 30° =16. N m

$1.6 N m$

$1.2 N m$

$1.4 N m$

$1.8 N m$

$τ = N I A B s in θ$
Here, $N = 25$ ,
$I = 10 A$ ,
$B = 0.9 T$ ,
$θ = 3 0^{\circ}$
$A = a^{2} = 12 \times 1 0^{- 2} \times 12 \times 1 0^{- 2}$
$144 \times 1 0^{- 4} m^{2}$
$∴ τ = 25 \times 10 \times = 144 \times 1 0^{- 4} \times 0.9 \times s in 3 0^{\circ}$
$= 16. N m$

Q. The magnitude of torque experienced by a square coil of side 12cm which consists of 25 turns and carries a current 10A suspended vertically and the normal to the plane of coil makes an angle of 30∘ with the direction of a uniform horizontal magnetic field of magnitude 0.9T is

1.6Nm

1.2Nm

1.4Nm

1.8Nm

τ=NIABsinθ Here, N=25, I=10A, B=0.9T, θ=30∘ A=a2=12×10−2×12×10−2 144×10−4m2 ∴τ=25×10×=144×10−4×0.9×sin30∘ =16.Nm

$1.6 N m$

$1.2 N m$

$1.4 N m$

$1.8 N m$

$τ = N I A B s in θ$
Here, $N = 25$ ,
$I = 10 A$ ,
$B = 0.9 T$ ,
$θ = 3 0^{\circ}$
$A = a^{2} = 12 \times 1 0^{- 2} \times 12 \times 1 0^{- 2}$
$144 \times 1 0^{- 4} m^{2}$
$∴ τ = 25 \times 10 \times = 144 \times 1 0^{- 4} \times 0.9 \times s in 3 0^{\circ}$
$= 16. N m$