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The intensity of magnetization of a bar magnet is $5.0 \times 10^{4} \mathrm{Am}^{-1}$. The magnetic length and the area of cross-section of the magnet are $12 \mathrm{cm}$ and $1 \mathrm{cm}^{2}$ respectively. The magnitude of magnetic moment of this bar magnet is (in SI unit)
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1668 Upvotes
Verified Answer
The correct answer is:
0.6
We know that Intensity of magnetisation
$I=\frac{M}{V}$
[where $M=$ magnetic moment, $V=$ volume]
So
$\begin{aligned}
M=I V &=5.0 \times 10^{4} \times \frac{12}{100} \times \frac{1}{(100)^{2}} \\
&=60 \times 10^{4} \times 10^{-6}=0.6 \mathrm{Am}^{2}
\end{aligned}$
$I=\frac{M}{V}$
[where $M=$ magnetic moment, $V=$ volume]
So
$\begin{aligned}
M=I V &=5.0 \times 10^{4} \times \frac{12}{100} \times \frac{1}{(100)^{2}} \\
&=60 \times 10^{4} \times 10^{-6}=0.6 \mathrm{Am}^{2}
\end{aligned}$
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