Question

The magnetic moment of a short bar magnet placed with its magnetic axis at ${30}^0$ to an external field of $900$G and experiences a torque of $0.02$N m is going to be

• A. $0.35A{m}^2$
• B. $0.44A{m}^2$
• C. $2.45A{m}^2$
• D. $1.5A{m}^2$

Answer 1

Answer: (B)

$0.44A{m}^2$

Here, $B=900Gauss=900\times {10}^{-4}T.=9\times {10}^{-2}T$
$T=0.02Nmand\theta ={30}^0$
$\therefore T=mBsin\theta$
$\Rightarrow 0.02=m\times 9\times {10}^{-2}\times sin{30}^0$
$m=\frac{0.02\times 2}{9\times {10}^{-2}}=0.44A{m}^2$.