Points $A$ and $B$ are situated along the extended axis of a $2 \ cm$ long bar magnet at distances $x$ and $2x$ $cm$ respectively from the nearer pole. The ratio of the magnetic field at $A$ and $B$ will be:

Two identical bar magnets of magnetic moment $M$ each, are placed along $X$ and $Y$-axes, respectively at a distance $d$ from the origin (as shown in the figure). The origin lies on the perpendicular bisector of the magnet placed on the $X$-axis and on the magnetic axis of the magnet placed on the $Y$-axis. If the magnitude of the total magnetic field at the origin is $B = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{d^3} \right]$, then the value of the constant $\alpha$ will be (given $d >> l$, where $l$ is the length of the bar magnets and the direction of $N$ to $S$ in the magnets is opposite with respect to each other).

Two small bar magnets are placed in a line with like poles facing each other at a certain distance $d$ apart. If the length of each magnet is negligible as compared to $d$,the force between them will be inversely proportional to

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

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

The work done in turning a magnet of magnetic moment $M$ by an angle $\theta$ from its initial position parallel to the magnetic meridian is: