Write the equation of torque acting on a bar magnet placed in a uniform magnetic field.

$A$ bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

Verify Ampere's law for the magnetic field of a point dipole with dipole moment $\vec{M} = M\hat{k}$. Take $C$ as the closed curve running clockwise along the quarter circle of radius $a$ and center at the origin in the first quadrant of the $x-z$ plane,closed by segments along the $x$ and $z$ axes.

$A$ bar magnet placed in a uniform magnetic field making an angle $\theta$ with the field experiences a torque. If the angle made by the magnet with the field is doubled,the torque experienced by the magnet increases by $41.4 \%$. The initial angle made by the magnet with the magnetic field is (in $^{\circ}$)

The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

$A$ small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06\, T$ experiences a torque of $0.018\, Nm$. The minimum work required to rotate it from its stable to unstable equilibrium position is