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Q)

A short bar magnet placed in a horizontal plane has its axis aligned along the magnet north - south direction . Null points are found on the axis of the of the magnet at $14\;cm$ from the center of the magnet . The earth magnetic field at the place is $0.36\;G$ and the cycle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null - point (14 cm) from the centre of the magnet?

$\begin{array}{1 1} \text{(i)magnitude : 0.54 G (ii) direction : along the direction of earth's field} \\ \text{(i)magnitude : 5.4 G (ii) direction: opposite to the direction of earth's field } \\\text{(i)magnitude : 0.54 G (ii) direction along the direction of earth's field} \\\text{ (i)magnitude : 0.54 G (ii) direction along the direction of earth's field} \end{array}$

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A)
Solution :
(At null points , field due to a magnet is equal and opposite to the horizontal component of earth magnetic field )
Given $d=14\;cm$
$\qquad= 0.14\;m$
$H= 0.6 \;G$
$B_1= \large\frac{\mu_0}{4 \pi} . \frac{2m}{d^3}$
But the magnetic field is equal to the horizontal component of earth's magnetic field .
(ie) $B_1=\large\frac{\mu_0}{4 \pi} . \frac{2m}{d^3}$
on the equatorial line of magnet at same distance (d) magnetic field due to the magnet.
$B_2 =\large\frac{\mu_0}{4 \pi} \frac{m}{d^3}$
$\qquad= \large\frac{B_1}{2}$
$\qquad= \large\frac{H}{2}$
The total magnetic field on equatorial line at this point is
$B= B_1+B_2$
$\quad= H+ \large\frac{H}{2}$
$\quad= \large\frac{3}{2}$$H \therefore B=\large\frac{3}{2}$$ \times 0.36$
$\qquad= 0.54\;G$
Hence the magnetic of earth's magnetic field is $0.54 \;G$
Answer : $\text{(i)magnitude : 0.54 G (ii) direction : along the direction of earth's field}$