(C) The magnetic moment $M$ of the bar magnet is given by the product of pole strength $m$ and the magnetic length $l$: $M = m \times l = 48 \, A \cdo

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(C) The magnetic moment $M$ of the bar magnet is given by the product of pole strength $m$ and the magnetic length $l$: $M = m \times l = 48 \, A \cdot m \times 0.25 \, m = 12 \, A \cdot m^2$. The torque $\tau$ acting on a magnetic dipole in a uniform magnetic field $B$ is given by $\tau = MB \sin \theta$. Given $M = 12 \, A \cdot m^2$,$B = 0.15 \, N \cdot A^{-1} \cdot m^{-1}$,and $\theta = 30^{\circ}$. $\tau = 12 \times 0.15 \times \sin(30^{\circ}) = 12 \times 0.15 \times 0.5 = 0.9 \, N \cdot m$.

Class 12 Physics Magnetism and Matter Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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The pole strength of a bar magnet is $48 \, A \cdot m$ and the distance between its poles is $25 \, cm$. The torque required to hold the magnet at an angle of $30^{\circ}$ with a uniform magnetic field of flux density $0.15 \, N \cdot A^{-1} \cdot m^{-1}$ is . . . . . . $N \cdot m$.

A
$12$
B
$18$
C
$0.9$
D
None of the above

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Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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The pole strength of a bar magnet is $48 \, A \cdot m$ and the distance between its poles is $25 \, cm$. The torque required to hold the magnet at an angle of $30^{\circ}$ with a uniform magnetic field of flux density $0.15 \, N \cdot A^{-1} \cdot m^{-1}$ is . . . . . . $N \cdot m$.

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$A$ magnet of magnetic moment $50 \hat{i} \text{ Am}^2$ is placed along the $x$-axis in a magnetic field $\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) \text{ T}$. The torque acting on the magnet is:

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