A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14\; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36\; G$ and the angle of $dip$ is zero.If the bar magnet is turned around by $180^o$, where will the new null points (in $cm$) be located?
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14\; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36\; G$ and the angle of $dip$ is zero.If the bar magnet is turned around by $180^o$, where will the new null points (in $cm$) be located?
The magnetic field on the axis of the magnet at a distance $d_{1}=14\, cm ,$ can be written as:
$B_{1}=\frac{\mu_{0} 2 M}{4 \pi\left(d_{1}\right)^{3}}=H\ldots .(1)$
Where, $M=$ Magnetic moment $\mu_{0}=$ Permeability of free space
$H=$ Horizontal component of the magnetic field at $d_{1}$
If the bar magnet is turned through $180^{\circ},$ then the neutral point will lie on the equatorial line. Hence, the magnetic field at a distance $d_{2},$ on the equatorial line of the magnet can be written
as:
$B_{2}=\frac{\mu_{0} M}{4 \pi\left(d_{2}\right)^{3}}=H\dots(ii)$
Equating equations $(i)$ and $(ii)$, we get:
$\frac{2}{\left(d_{1}\right)^{3}}=\frac{1}{\left(d_{2}\right)^{3}}$
$\left(\frac{d_{2}}{d_{1}}\right)^{3}=\frac{1}{2}$
$\therefore d_{2}=d_{1} \times\left(\frac{1}{2}\right)^{1 / 3}$
$=14 \times 0.794=11.1 \,cm$
The new null points will be located $11.1 \;cm$ on the normal bisector.
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