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(B) Let the two magnets be $M_1$ and $M_2$, each of length $\ell$ and pole strength $m$. The magnetic moment of each magnet is $\mu = m \ell$.Since th

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$\sqrt{2} m \ell$

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(B) Let the two magnets be $M_1$ and $M_2$, each of length $\ell$ and pole strength $m$. The magnetic moment of each magnet is $\mu = m \ell$.Since they are placed at right angles with the north pole of one touching the south pole of the other, the magnetic moments $\vec{\mu}_1$ and $\vec{\mu}_2$ are perpendicular to each other.The resultant magnetic moment $\vec{\mu}_{res}$ is given by the vector sum: $\vec{\mu}_{res} = \vec{\mu}_1 + \vec{\mu}_2$.The magnitude of the resultant magnetic moment is $\mu_{res} = \sqrt{\mu_1^2 + \mu_2^2}$.Substituting $\mu_1 = \mu_2 = m \ell$, we get:$\mu_{res} = \sqrt{(m \ell)^2 + (m \ell)^2} = \sqrt{2 (m \ell)^2} = \sqrt{2} m \ell$.Thus, the magnitude of the resultant magnetic moment is $\sqrt{2} m \ell$.

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 $\ell$. The magnitude of the resultant magnetic moment of the system is $[m = \text{pole strength of the magnet}]$.

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