Question

A small bar magnet experiences a torque of $0.016Nm$ when placed with its axis at $30^{\circ }$ with an external field of $0.04T$. If the bar magnet is replaced by a solenoid of cross-sectional area of $1c{m}^2$ and $1000$ turns but having the same magnetic moment as that of bar magnet, then the current flowing through the solenoid is

• A. 2 A
• B. 4 A
• C. 6 A
• D. 8 A

Answer 1

Answer: (D)

Given,
torque on a bar magnet, $\tau =0.016Nm$,
$\theta =30^{\circ }$
and magnetic field, $B=0.04T$
Since, $\tau =mB\sin \theta$
$\therefore 0.016=m\times 0.04\times \sin 30^{\circ }$
$\Rightarrow m=\frac{2\times 0.016}{0.04}$
$m=0.8A{m}^2$
For solenoid, area of cross-section,
$A=1c{m}^2=10^{-4}{m}^2$
Number of turns in solenoid,
$N=1000$
$\therefore$ magnetic moment of solenoid,
$m=NIA$
$\Rightarrow I=\frac{m}{NA}$
$=\frac{0.8}{1000\times 10^{-4}}$ [From Eq.(i)]
$=8A$