A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines make an angle of 30° with the normal of the -coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is

Question

A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines make an angle of 30° with the normal of the -coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is (A) 33 N m (B) 3.3 N m (C) 3.3 × 10-2 N m (D) 3.3 × 10-4 N m

Tardigrade Admin · Accepted Answer

N=70, r=5 cm=5 × 10-2 m, I=8 A B=1.5 T, θ = 30° The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil, ∴ τ= NIAB sin θ = NI π r2B sin 30° =70× 8× 3.14 × (5 × 10-2)2 × 1.5 × (1/2)= 3.297 N m ≈ 3.3 N m

$33 N m$

$3.3 N m$

$3.3 \times 1 0^{- 2} N m$

$3.3 \times 1 0^{- 4} N m$

33Nm

3.3Nm

3.3×10−2Nm

3.3×10−4Nm

N=70, r=5cm=5×10−2m, I=8A B=1.5T, θ=30∘ The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil, ∴τ=NIABsinθ=NIπr2Bsin30∘ =70×8×3.14×(5×10−2)2×1.5×21​=3.297Nm ≈3.3Nm

$33 N m$

$3.3 N m$

$3.3 \times 1 0^{- 2} N m$

$3.3 \times 1 0^{- 4} N m$