$A$ 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 $\text{ J}$)

$A$ 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:

If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field, then the torque acting on the magnet is:

$A$ 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 $= B_H$)

$A$ short bar magnet of magnetic moment $0.21 \ A \cdot m^2$ is placed with its axis perpendicular to the direction of the horizontal component of the earth's magnetic field. The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at $45^{\circ}$ with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field $= 4.2 \times 10^{-5} \ T$). (in $cm$)

$A$ magnet of magnetic moment $M$ is lying in a uniform magnetic field $B$. $W_1$ is the work done in turning it from $0^o$ to $60^o$ and $W_2$ is the work done in turning it from $30^o$ to $90^o$. Then: