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Two identical thin bar magnets, each of length \(l\) and pole strength \(m\), are placed at right angle to each other with north pole of one touching south pole of the other. The magnetic moment of the system is
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The correct answer is:
\(\sqrt{2} / m\)
The given situations of two bar magnets is shown in the following figure.
Magnetic dipole moment of each bar magnet
\(M_1=M_2=M=m l\)
\(M_1\) and \(M_2\) are perpendicular to each other as shown in the figure.
\(\therefore\) Net magnetic moment of the system,
\(M=\sqrt{M_1^2+M_2^2+2 M_1 M_2 \cos 90^{\circ}}\)
\(\begin{aligned}
& =\sqrt{M_1^2+M_2^2}=\sqrt{M^2+M^2} \\
& =M \sqrt{2}=m l \sqrt{2}=\sqrt{2} m l
\end{aligned}\)
Magnetic dipole moment of each bar magnet
\(M_1=M_2=M=m l\)
\(M_1\) and \(M_2\) are perpendicular to each other as shown in the figure.
\(\therefore\) Net magnetic moment of the system,
\(M=\sqrt{M_1^2+M_2^2+2 M_1 M_2 \cos 90^{\circ}}\)
\(\begin{aligned}
& =\sqrt{M_1^2+M_2^2}=\sqrt{M^2+M^2} \\
& =M \sqrt{2}=m l \sqrt{2}=\sqrt{2} m l
\end{aligned}\)
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