A short bar magnet is placed in a horizontal plane with its axis in the magnetic meridian. Null points are found on its equatorial line (i.e., its normal bisector) at 12.5 cm from the centre of the magnet. The earth's magnetic field at the place is 0.38 gauss, and the angle of dip is zero.
(a) What is the total magnetic field at points on the axis of the magnet located at the same distance (12.5 cm) as the null points from the centre?
(b) Locate the null points when the bar is turned around by 180°. Assume that the length of the magnet is negligible compared to the distance of the null points from the centre of the magnet.

 
                            Fig.(a) 

(a)
From fig. (a), it is clear that null points are obtained on the normal bisector when the magnet’s north and south poles face magnetic north and south respectively.
Magnetic field on the normal bisector at a distance r from the centre is given by 

               $\overrightarrow{{\mathrm{B}}_{\mathrm{e}}} = -\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\overrightarrow{\mathrm{M}}}{{\mathrm{r}}^3},$  

provided r is much greater than the length of the magnet. [The above equation is strictly true only for a point dipole]. 

At a null point, this field is balanced by the earth's field.
Earth's magnetic field,B = 0.38 G
Angle of dip = 0^o

So,          ${\mathrm{B}}_{\mathrm{e}} = {\mathrm{B}}_{\mathrm{H}}$ 

           $\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\mathrm{r}}^3} = 0.38 \times {10}^4$          ...(1) 

Since, angle of dip = 0, therefore, B_v = 0 and the horizontal component of the earth’s field equals the field itself. 

Next, magnetic field due to a magnet on its axis at point distant r from the centre is given by

               ${\mathrm{B}}_{\mathrm{a}} = \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\mathrm{r}}^3}$                    ...(2) 

provided r is much greater than the length of the magnet. (The above equation is strictly true only for a point dipole).
From fig. (a), it is clear that on the axis, this field adds up to the earth’s field. 

Thus, the total field at a point on the axis has a magnitude equal to B_a + B_H

i.e.,  $\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\mathrm{r}}^3}+0.38 \times {10}^{-4}$              ...(3) 

and direction along $\overrightarrow{\mathrm{M}}$ (magnetic moment) [which is parallel to the earth's field in case of (a)]. 

Thus, for the same distance on the axis as the distance of the null point, the total field is found using equations (1) and (3).
That is,
B = 2 x 0.38 x 10^–4 + 0.38 x 10^–4
   = 3 x 0.38 x 10^–4 
   = 1.14 x 10^–4 T 

This field is directed along $\overrightarrow{\mathrm{M}}.$ 

Note: We did not require the given value of 12.5 cm for the nullpoint distance, except in so far that this was assumed to be much greater than the length of the magnet. 

(b) When the bar is turned around by 180°, the magnet’s north and south poles face magnetic south and north respectively i.e., in this case, $\overrightarrow{\mathrm{M}}$is antiparallel to the earth's field. 

 

                      Fig. (b)

From fig. (b), it is clear that the nullpoints now lie on the axis of the magnet at a distance r’ given by B_a’ = B_H 

        $\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{M}}{\mathrm{r}{\text{'}}^3} = 0.38 \times {10}^{-4}$             ...(4) 

Comparing equations (4) and (1), we get 

                       $$\begin{gathered} \frac{2}{\mathrm{r}{\text{'}}^3} = \frac{1}{{\mathrm{r}}^3} \\ \mathrm{r}{\text{'}}^3 = 2{\mathrm{r}}^3 \\ \mathrm{r}\text{'} = (2{)}^{1/3 }\mathrm{r} \end{gathered}$$
Therefore,
                  $\mathrm{r}= 12.5 \mathrm{cm}, \mathrm{r}\text{'} = 15.7 \mathrm{cm}$