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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. What is the total magnetic field (in $G$) on the normal bisector of the magnet at the same distance as the null-point (i.e., $14 \;cm$) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)
$0.72$
$0.18$
$0.36$
$0.54$
Solution
Earth's magnetic field at the given place, $H=0.36\, G$ The magnetic field at a distance $d$, on the axis of the magnet is given as:
$B_{1}=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}=H\dots(i)$
Where, $\mu_{0}=$ Permeability of free space
$M=$ Magnetic moment The magnetic field at the same distance $d$, on the equatorial line of the magnet is given as:
$B_{2}=\frac{\mu_{0} M}{4 \pi d^{3}}=\frac{H}{2}$ [Using equation $(i)$]
Total magnetic field, $B=B_{1}+B_{2}$ $=H+\frac{H}{2}$
$=0.36+0.18=0.54 \,G$
Hence, the magnetic field is $0.54 \,G$ in the direction of earth's magnetic field.
