A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm. The magnetic moment of the magnet is

Question

A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm. The magnetic moment of the magnet is (A) 4 Am 2 (B) 0.5 Am 2 (C) 2 Am 2 (D) 0.40 Am 2

Tardigrade Admin · Accepted Answer

The magnetic moment of the magnet τ=m B sin θ, θ=30° hence, sin θ=1 / 2 (∵ 1 G =10-4 T ) Thus, 0.016=m ×(800 × 10-4 T ) ×(1 / 2) ⇒ m=160 × 2 / 800=0.40 Am 2

Q. A short bar magnet placed with its axis at $3 0^{\circ}$ with an external field of $800 G$ experiences a torque of $0.016 N m$ . The magnetic moment of the magnet is

$4 A m^{2}$

$0.5 A m^{2}$

$2 A m^{2}$

$0.40 A m^{2}$

The magnetic moment of the magnet
$τ = m B sin θ, θ = 3 0^{\circ}$
hence, $sin θ = 1/2 (∵ 1 G = 1 0^{- 4} T)$
Thus, $0.016 = m \times (800 \times 1 0^{- 4} T) \times (1/2)$
$\Rightarrow m = 160 \times 2/800 = 0.40 A m^{2}$

Q. A short bar magnet placed with its axis at 30∘ with an external field of 800G experiences a torque of 0.016Nm. The magnetic moment of the magnet is

4Am2

0.5Am2

2Am2

0.40Am2