$A$ 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 $\text{ J}$)

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

$A$ bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it,then the angle by which it is to be rotated is....$^o$

$A$ 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:

Two short bar magnets have magnetic moments $1.2 \text{ Am}^2$ and $1.0 \text{ Am}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres, such that their north poles point towards the geographic south. They share a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. Calculate the resultant horizontal magnetic induction at the midpoint of the line joining their centres. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2$)

$A$ 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: