The magnetic field due to a bar magnet at a distance $R$ from the center of the magnet (where $R$ is much larger than the length of the magnet) is proportional to:

Torques $\tau_1$ and $\tau_2$ are required for a magnetic needle to remain perpendicular to the magnetic fields of $B_1$ and $B_2$ at two different places. The ratio of $B_1: B_2$ is equal to

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

$A$ magnetic needle of magnetic moment $6 \times 10^{-2} \text{ A m}^2$ and moment of inertia $9.6 \times 10^{-5} \text{ kg m}^2$ performs simple harmonic motion in a magnetic field of $0.01 \text{ T}$. The time taken to complete $10$ oscillations is (Take $\pi = 3.14$): (in $\text{ s}$)

The midpoints of two small magnetic dipoles of length $d$ in end-on positions are separated by a distance $x$ $(x >> d)$. The force between them is proportional to $x^{-n}$,where $n$ is:

The torque acting on a magnetic dipole placed in a uniform magnetic field is zero,when the angle between the dipole axis and the magnetic field is