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A small bar magnet experiences a torque of $0.016 \mathrm{Nm}$ when placed with its axis at $30^{\circ}$ with an external field of $0.04 \mathrm{~T}$. If the bar magnet is replaced by a solenoid of cross-sectional area of $1 \mathrm{~cm}^2$ and 1000 turns but having the same magnetic moment as that of bar magnet, then the current flowing through the solenoid is
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Verified Answer
The correct answer is:
$8 \mathrm{~A}$
Given,
torque on a bar magnet, $\tau=0.016 \mathrm{Nm}$, $\theta=30^{\circ}$ and magnetic field, $B=0.04 \mathrm{~T}$
Since, $\tau=m \mathrm{~B} \sin \theta$
$\begin{array}{l}
\therefore 0.016=m \times 0.04 \times \sin 30^{\circ} \\
\Rightarrow \quad m=\frac{2 \times 0.016}{0.04} \\
m=0.8 \mathrm{Am}^2
\end{array}$
For solenoid, area of cross-section,
$A=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2$
Number of turns in solenoid,
$N=1000$
$\therefore$ magnetic moment of solenoid,
$\begin{aligned}
m & =N I A \\
\Rightarrow \quad I & =\frac{m}{N A} \\
& =\frac{0.8}{1000 \times 10^{-4}} \quad \text { [From Eq. (i)] } \\
& =8 \mathrm{~A}
\end{aligned}$
torque on a bar magnet, $\tau=0.016 \mathrm{Nm}$, $\theta=30^{\circ}$ and magnetic field, $B=0.04 \mathrm{~T}$
Since, $\tau=m \mathrm{~B} \sin \theta$
$\begin{array}{l}
\therefore 0.016=m \times 0.04 \times \sin 30^{\circ} \\
\Rightarrow \quad m=\frac{2 \times 0.016}{0.04} \\
m=0.8 \mathrm{Am}^2
\end{array}$
For solenoid, area of cross-section,
$A=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2$
Number of turns in solenoid,
$N=1000$
$\therefore$ magnetic moment of solenoid,
$\begin{aligned}
m & =N I A \\
\Rightarrow \quad I & =\frac{m}{N A} \\
& =\frac{0.8}{1000 \times 10^{-4}} \quad \text { [From Eq. (i)] } \\
& =8 \mathrm{~A}
\end{aligned}$
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