Two identical short bar magnets,each having magnetic moment $M$,are placed a distance of $2d$ apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is

$A$ 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....$^o$

$A$ bar magnet having a magnetic moment of $2 \times 10^4 \, JT^{-1}$ is free to rotate in a horizontal plane. $A$ horizontal magnetic field $B = 6 \times 10^{-4} \, T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is.....$J$

The effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows:

$A$ 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 $= B_H$)

$A$ bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along the positive $X$-axis. Likewise, a second bar magnet of magnetic moment $M$ is placed at a distance $2D$ on the positive $Y$-axis and perpendicular to it as shown in the figure. The magnitude of the magnetic field at the origin is $|\vec{B}| = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{D^3} \right]$. The value of $\alpha$ must be (Assume $D \gg l$, where $l$ is the length of the magnets).