Question

The magnetic moment of a short bar magnet placed with its magnetic axis at $30^{\circ}$ to an external field of $900\, G$ and experiences a torque of $0.02\, N\, m$ is

• A. $0.35 \,A \,m^2$
• B. $0.44 \,A \,m^2$
• C. $2.45 \,A \,m^2$
• D. $1.5 \,A \,m^2$

Answer 1

Answer: (B)

$0.44 \,A \,m^2$

Answer 2

Here, $B = 900$ Gauss $= 900 \times 10^{-4} \,T$. $= 9 \times 10^{-2} \,T$ $\tau = 0.02\,N\,m$ and $\theta =30^{\circ}$ $\because\quad\tau = mB \,sin\,\theta$ $\Rightarrow\quad0.02 = m \times 9 \times 10^{-2} \times sin\, 30^{\circ}$ $0.02 = 9 \times 10^{-2} \times \frac{1}{2} \times m$ $m = \frac{0.02 \times 2}{9 \times 10^{-2}} = 0.44\,A\,m^{2}$.