Two identical thin bar magnets are placed mutually at right angles such that the north pole of one touches…
Two identical thin bar magnets are placed mutually at right angles such that the
north pole of one touches the south pole of the other. The length of each bar
magnet is ' $\ell^{\prime}$. The magnitude of resultant magnetic moment of the system is
$[\mathrm{m}=$ pole strength of the pole of magnet $]$
$2 \mathrm{~m} \ell$
$\sqrt{2} \mathrm{~m} \ell$
$\mathrm{m} \ell$
$\sqrt{3} \mathrm{~m} \ell$
Solution
Given that,
Length of bar magnet $=\mathrm{L}$
Pole strength $=\mathrm{m}$
Let the pole strength of the system magnets be $m$ and length $N_{2} S_{1}$
But,
$\begin{array}{l}
\mathrm{N}_{2} \mathrm{~S}_{1}=\sqrt{\left(\mathrm{N}_{1} \mathrm{~S}_{1}\right)^{2}+\left(\mathrm{N}_{2} \mathrm{~S}_{2}\right)^{2}} \\
\mathrm{~N}_{1} \mathrm{~S}_{1}=\mathrm{N}_{2} \mathrm{~S}_{2}=\mathrm{L} \\
\mathrm{N}_{2} \mathrm{~S}_{1}=\sqrt{\mathrm{L}^{2}+\mathrm{L}^{2}} \\
\mathrm{~N}_{2} \mathrm{~S}_{1}=\mathrm{L} \sqrt{2}
\end{array}$
Magnetic moment of the system
$\begin{array}{l}
\mathrm{M}=\mathrm{m} \times \mathrm{N}_{2} \mathrm{~S}_{1} \\
\mathrm{M}=\mathrm{m} \times \mathrm{L} \sqrt{2} \\
\mathrm{M}=\mathrm{mL} \sqrt{2}
\end{array}$
Hence, the Magnetic moment of the system is $\mathrm{mL} \sqrt{2}$
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