The final torque on a coil having magnetic moment 25 A m2 in a 5 T uniform external magnetic field, if the coil rotates through an angle of 60° under the influence of the magnetic field is

Question

The final torque on a coil having magnetic moment 25 A m2 in a 5 T uniform external magnetic field, if the coil rotates through an angle of 60° under the influence of the magnetic field is (A) 216.5 N m (B) 108.25 N m (C) 102.5 N m (D) 258.1 N m

Tardigrade Admin · Accepted Answer

| vecτ|=| vecm× vecB |=mB sin θ Here, m=25 A m2 ; θ=60°; B=5 T ∴ τ=25 × 5 × sin 60° or τ =125 × (√3/2) =108.25 N m

Q. The final torque on a coil having magnetic moment $25 A m^{2}$ in a $5 T$ uniform external magnetic field, if the coil rotates through an angle of $6 0^{\circ}$ under the influence of the magnetic field is

$216.5 N m$

$108.25 N m$

$102.5 N m$

$258.1 N m$

$∣ τ ∣ = ∣ ∣ m \times B ∣ ∣ = m B s in θ$
Here, $m = 25 A m^{2}; θ = 6 0^{\circ}; B = 5 T$
$∴ τ = 25 \times 5 \times s in 6 0^{\circ}$ or $τ = 125 \times \frac{3}{2}$
$= 108.25 N m$

Q. The final torque on a coil having magnetic moment 25Am2 in a 5T uniform external magnetic field, if the coil rotates through an angle of 60∘ under the influence of the magnetic field is

216.5Nm

108.25 Nm

102.5Nm

258.1Nm