(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

Easy

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

In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?

EasyGUJCET 2009

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MediumMHT CET 2007

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$A$ bar magnet of length $3 \, cm$ has points $A$ and $B$ along its axis at distances of $24 \, cm$ and $48 \, cm$ on the opposite sides. The ratio of the magnetic fields at these points is

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$A$ small bar magnet of magnetic moment $M$ is placed in a uniform magnetic field $H$. If the magnet makes an angle of $30^{\circ}$ with the field,the torque acting on the magnet is:

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The ratio of magnetic intensities for an axial point and a point on the broad side-on (equatorial) position at an equal distance $d$ from the centre of a short bar magnet is:

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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

In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?

At a point on the equatorial line (right bisector) of a magnetic dipole,the magnetic:

$A$ bar magnet of length $3 \, cm$ has points $A$ and $B$ along its axis at distances of $24 \, cm$ and $48 \, cm$ on the opposite sides. The ratio of the magnetic fields at these points is

$A$ small bar magnet of magnetic moment $M$ is placed in a uniform magnetic field $H$. If the magnet makes an angle of $30^{\circ}$ with the field,the torque acting on the magnet is:

The ratio of magnetic intensities for an axial point and a point on the broad side-on (equatorial) position at an equal distance $d$ from the centre of a short bar magnet is:

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