(A) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{m} \cdot \vec{B} = -mB \cos \theta$.For $\theta = 0^{\circ}$,

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Stable equilibrium at $\theta = 0^{\circ}$,Unstable equilibrium at $\theta = 180^{\circ}$

(A) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{m} \cdot \vec{B} = -mB \cos \theta$.For $\theta = 0^{\circ}$,$U = -mB \cos(0^{\circ}) = -mB$,which is the minimum potential energy. Therefore,the magnet is in a state of stable equilibrium.For $\theta = 180^{\circ}$,$U = -mB \cos(180^{\circ}) = +mB$,which is the maximum potential energy. Therefore,the magnet is in a state of unstable equilibrium.

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

Medium

Give the stability position of a bar magnet for $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$.

A
Stable equilibrium at $\theta = 0^{\circ}$,Unstable equilibrium at $\theta = 180^{\circ}$
B
Unstable equilibrium at $\theta = 0^{\circ}$,Stable equilibrium at $\theta = 180^{\circ}$
C
Stable equilibrium at both positions
D
Unstable equilibrium at both positions

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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

Two identical bar magnets,each of magnetic moment $M$,are kept perpendicular to each other at a certain distance. The magnetic induction at a point that is at the same distance $d$ from the center of both magnets is: (where $\mu_{0}$ is the permeability of free space)

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Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on (axial) position is $9 \ G$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on (equatorial) position (in $G$)?

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$A$ magnet of total magnetic moment $10^{-2} \hat{i} \text{ A} \cdot \text{m}^2$ is placed in a time-varying magnetic field $\vec{B} = B \cos(\omega t) \hat{i}$, where $B = 1 \text{ T}$ and $\omega = 0.125 \text{ rad/s}$. The work done to reverse the direction of the magnetic moment at $t = 1 \text{ s}$ is: (in $\text{ J}$)

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Points $A$ and $B$ are situated perpendicular to the axis of a $2 \ cm$ long bar magnet at large distances $X$ and $3X$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to:

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$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \ T$ experiences a torque of magnitude $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment of the magnet will be . . . . . . $J \ T^{-1}$.

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Give the stability position of a bar magnet for $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$.

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Two identical bar magnets,each of magnetic moment $M$,are kept perpendicular to each other at a certain distance. The magnetic induction at a point that is at the same distance $d$ from the center of both magnets is: (where $\mu_{0}$ is the permeability of free space)

Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on (axial) position is $9 \ G$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on (equatorial) position (in $G$)?

Points $A$ and $B$ are situated perpendicular to the axis of a $2 \ cm$ long bar magnet at large distances $X$ and $3X$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to:

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \ T$ experiences a torque of magnitude $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment of the magnet will be . . . . . . $J \ T^{-1}$.

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