A short bar magnet placed in a horizontal plane has its axis alig… - Vidyadip

Question

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-sourth 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 on the normal bisector of the magnet at the same distance as the null points (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).

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Answer

As null points are on the axis of the magnet, therefore,
$B_1=\frac{\mu_0}{4 \pi} \frac{2 M}{d^3}=H$
On the equatorial line of magnet at same direction (d), field due to the magnet is
$B_2=\frac{\mu_0}{4 \pi} \frac{M}{d^3}=\frac{B_1}{2}=\frac{H}{2}$
$\therefore$ Total magnetic field at this point on equatorial line
$B = H + B _2= H +\frac{ H }{2}$
$=\frac{3}{2} H =\frac{3}{2} \times 0.36=0.54 G$

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