Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on (axial) position is $9 \ G$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on (equatorial) position (in $G$)?

$A$ magnetic dipole is placed in a uniform magnetic field of intensity $B$,oriented along the direction of the field. If the magnetic dipole moment is $M$,then the maximum work an external agent can perform in rotating the dipole will be

$A$ bar magnet,when placed at an angle of $30^o$ to the direction of a magnetic field of induction $5 \times 10^{-2} \,T$,experiences a torque of $25 \times 10^{-6} \,N-m$. If the length of the magnet is $5 \,cm$,what is its pole strength?

$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

If the dipole moment of a short bar magnet is $1.25 \ A-m^2$,find the magnetic field on its axis at a distance of $0.5 \ m$ from the centre of the magnet.

$A$ 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).