$A$ 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 $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

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:

Two magnetic dipoles $X$ and $Y$ are placed at a separation $d$,with their axes perpendicular to each other. The dipole moment of $Y$ is twice that of $X$. $A$ particle of charge $q$ is passing through their mid-point $P$,at an angle $\theta = 45^\circ$ with the horizontal line as shown in the figure. What would be the magnitude of the force on the particle at that instant? ($d$ is much larger than the dimensions of the dipole)

$A$ bar magnet of magnetic moment $1.5 \, J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment perpendicular to the field direction? (in $J$)

$A$ magnet suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by $19 \%$. By doing this,the time period of the magnet approximately

$A$ 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: