Question

The magnetic moment of a short bar magnet placed with its magnetic axis at ${30}^o$ 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 $m^2$

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