Two bar magnets having same geometry with magnetic moments $M$ and $2 M$, are firstly placed in such a way that their similar poles are same side then its time period of oscillation is $T_{1}$. Now the polarity of one of the magnet is reversed then time period of oscillation is $T_{2},$ then
$T_{1} < T_{2}$
$T_{1}=T_{2}$
$T_{1}>T_{2}$
$T_{2}=\infty$
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 $4×10^{-5}\, T$ to a direction $ 60° $ from the field will be.....$ J$
Three identical bar magnets each of magnetic moment $M$ are placed in the form of an equilateral triangle as shown. The net magnetic moment of the system is
A bar magnet of length $'l'$ and magnetic dipole moment $'M'$ is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
Two identical short bar magnets, each having magnetic moment of $10\, Am^2$, are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is $0.2 \,m$ , the resultant magnetic induction at a point midway between them is$({\mu _0} = 4\pi \times {10^{ - 7}}\,H{m^{ - 1}})$
The magnetic field to a small magnetic dipole of magnetic moment $M$, at distance $ r$ from the centre on the equatorial line is given by (in $M.K.S. $ system)