What is the magnitude of the equatorial magnetic field due to a bar magnet of length $5.0 \ cm$ at a distance $75 \ cm$ from its midpoint? The magnetic moment of the bar magnet is $0.75 \ A \ m^2$.

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 . . . . . . $N \cdot m$.

$A$ 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 equal to $4.5 \times 10^{-2} \; J$. What is the magnitude of magnetic moment (in $J \; T^{-1}$) of the magnet?

$A$ magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^{\circ}$. One of the fields has a magnitude of $15 \, mT$. The dipole attains stable equilibrium at an angle of $30^{\circ}$ with this field. The magnitude of the other field (in $mT$) is close to:

$A$ negative charge is given to a nonconducting loop and the loop is rotated in the plane of paper about its centre as shown in the figure. The magnetic field produced by the ring affects a small magnet placed above the ring in the same plane:

$A$ donut-shaped permanent magnet (magnetization parallel to the axis) can slide frictionlessly on a vertical rod. Treat the magnets as dipoles with mass $m_d$ and dipole moment $M$. When we put two back-to-back magnets on the rod,the upper one will float. At what height $z$ does it float?