(A) When a magnet is displaced by a very small angle $\theta$, the restoring torque acting on the magnet is given by $\tau = -M B_H \sin \theta$. The

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(A) When a magnet is displaced by a very small angle $\theta$, the restoring torque acting on the magnet is given by $\tau = -M B_H \sin \theta$. The negative sign indicates the restoring nature of the torque. Since $\tau = I \alpha$, where $I$ is the moment of inertia and $\alpha$ is the angular acceleration, we have $I \alpha = -M B_H \sin \theta$. For small angular displacements, $\sin \theta \approx \theta$. Therefore, $I \alpha = -M B_H \theta$. The magnitude of the angular acceleration is $\alpha = \frac{M B_H \theta}{I}$.

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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$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$, the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

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A
$\frac{M B_H \theta}{I}$
B
$\frac{I B_H \theta}{M}$
C
$\frac{M \theta}{I B_H}$
D
$\frac{I \theta}{M B_H}$

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Magnetism and Matter

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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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$A$ short bar magnet of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

MediumAP EAMCET 2025

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Points $A$ and $B$ are situated along the extended axis of a $2 \ cm$ long bar magnet at distances $x$ and $2x$ $cm$ respectively from the nearer pole. The ratio of the magnetic field at $A$ and $B$ will be:

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$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

Medium

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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 suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by $19 \%$. By doing this,the time period of the magnet approximately

MediumAP EAMCET 2024

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$A$ short bar magnet of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

Points $A$ and $B$ are situated along the extended axis of a $2 \ cm$ long bar magnet at distances $x$ and $2x$ $cm$ respectively from the nearer pole. The ratio of the magnetic field at $A$ and $B$ will be:

$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

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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