(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 n

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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

Difficult

$A$ 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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MediumJEE MAIN 2019

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Medium

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Medium

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$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$,find the magnetic induction at point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

The work done in rotating a bar magnet from its equilibrium position by $60^{\circ}$ in a magnetic field is $W$. What is the torque required to hold it in this position?

$A$ bar magnet of magnetic moment $1.5 \, A \cdot m^2$ lies aligned with the direction of a uniform magnetic field of $2 \, T$. What is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction?

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