A short bar magnet with its north pole facing north forms a neutral point at $P$ in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane, the net magnetic induction at $P$ is (Horizontal component of earth’s magnetic field = ${B_H}$)
$0$
$2 BH$
$\frac{{\sqrt 5 }}{2}{B_H}$
$\sqrt 5 \,{B_H}$
What is pole strength of a bar magnet ? Show the magnetic dipole moment in terms of pole strength.
A small bar magnet has a magnetic moment $1.2 \,A-m^2$. The magnetic field at a distance $ 0.1\, m $ on its axis will be : ($\mu_0 = 4\pi \times 10^{-7} \,T-m/A$)
Some equipotential surfaces of the magnetic scalar potential are shown in the figure. Magnetic field at a point in the region is
Two identical bar magnets with a length $10 \,cm $ and weight $50 \,gm$-weight are arranged freely with their like poles facing in a inverted vertical glass tube. The upper magnet hangs in the air above the lower one so that the distance between the nearest pole of the magnet is $3\,mm.$ Pole strength of the poles of each magnet will be.......$ amp × m$
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)