(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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Magnetism and Matter

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

If a magnet of length $10 \, cm$ and pole strength $40 \, A-m$ is placed at an angle of $45^\circ$ in a uniform magnetic field of intensity $2 \times 10^{-4} \, T$,the torque acting on it is:

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$A$ bar magnet with a magnetic moment $5.0\,Am^2$ is placed in a parallel position relative to a magnetic field of $0.4\,T$. The amount of work done in turning the magnet from a parallel to an antiparallel position relative to the field direction is $.........\,J$.

MediumJEE MAIN 2023

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$A$ short bar magnet placed with its axis at $45^{\circ}$ with a uniform external magnetic field of $28.3 \times 10^{-3} \,T$ experiences a torque of magnitude equal to $3.6 \times 10^{-5} \,J$. The magnitude of the magnetic moment of the magnet is nearly:

EasyAP EAMCET 2022

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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 $= B_H$)

DifficultTS EAMCET 2007

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The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is:

MediumAIPMT 1995

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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 . . . . . . $N \cdot m$.

Similar Questions

If a magnet of length $10 \, cm$ and pole strength $40 \, A-m$ is placed at an angle of $45^\circ$ in a uniform magnetic field of intensity $2 \times 10^{-4} \, T$,the torque acting on it is:

$A$ bar magnet with a magnetic moment $5.0\,Am^2$ is placed in a parallel position relative to a magnetic field of $0.4\,T$. The amount of work done in turning the magnet from a parallel to an antiparallel position relative to the field direction is $.........\,J$.

$A$ short bar magnet placed with its axis at $45^{\circ}$ with a uniform external magnetic field of $28.3 \times 10^{-3} \,T$ experiences a torque of magnitude equal to $3.6 \times 10^{-5} \,J$. The magnitude of the magnetic moment of the magnet is nearly:

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is:

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