Question

A small bar magnet experiences a torque of $0.016 Nm$ when placed with its axis at $30^{\circ}$ with an external field of $0.04 \,T$. If the bar magnet is replaced by a solenoid of cross-sectional area of $1 cm ^{2}$ 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, $\tau=0.016 Nm$,
$\theta=30^{\circ}$
and magnetic field, $B=0.04 T$
Since, $\tau=m B \sin \theta$
$\therefore 0.016=m \times 0.04 \times \sin 30^{\circ}$
$\Rightarrow m=\frac{2 \times 0.016}{0.04} $
$ m=0.8 Am ^{2}$
For solenoid, area of cross-section,
$A=1 cm ^{2}=10^{-4} m ^{2}$
Number of turns in solenoid,
$N=1000$
$\therefore $ magnetic moment of solenoid,
$m=N I A$
$\Rightarrow I=\frac{m}{N A}$
$=\frac{0.8}{1000 \times 10^{-4}} $ [From Eq.(i)]
$=8 A$