The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

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

The magnetic moment of a current-carrying loop is $2.1 \times 10^{-25} \text{ A m}^2$. The magnetic field at a point on its axis at a distance of $1 \text{ Å}$ is:

Two short bar magnets $A$ and $B$ (having magnetic moments $M_{1}$ and $M_{2}$ respectively) are kept one above the other with their magnetic axes perpendicular to each other. If their resultant magnetic field at a point on the axis of magnet $A$ is inclined at $45^{\circ}$ with the axis of magnet $A$,then the ratio of magnetic moments $\frac{M_{2}}{M_{1}}$ is $[\tan 45^{\circ} = 1]$.

$A$ magnetic dipole of moment $2.5 \text{ A m}^2$ is free to rotate about a vertical axis passing through its centre. It is released from the East-West direction. What is its kinetic energy at the moment it takes the North-South position (in $\mu\text{J}$)? (Given: $B_H = 3 \times 10^{-5} \text{ T}$)

Write the equation of torque on a magnetic needle placed in a uniform magnetic field and derive the expression for its time period $T = 2\pi \sqrt{\frac{I}{mB}}$.