Ncert Exercises & Solutions

5.14 If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points 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{μ_{0} 2 M}{4 π {(d_{1})}^{3}} = H . . . (i)

Where M = Magnetic moment

μ_{0}

= Permeability of free space, H= Horizontal component
of the magnetic field at

d_{1}

If the bar magnet is turned through 180

°

,
then the neutral point will lie on the equilateral line.
Hence, the magnetic field at a distance

d_{2}

, on the equatorial l in of the magnet can be written as

B_{2} = \frac{μ_{0} M}{4 π {(d_{2})}^{3}} = H . . . (ii)

Equating equations (i) and (ii), we get :

\frac{2}{{(d_{1})}^{3}} = \frac{1}{{(d_{2})}^{3}}

\begin{array}{l} \Rightarrow {(\frac{d_{2}}{d_{1}})}^{3} = \frac{1}{2} \\ \Rightarrow d_{2} = d_{1} \times {(\frac{1}{2})}^{\frac{1}{3}} \\ \Rightarrow d_{2} = 14 \times 0 .794 = 11 .1 cm \end{array}

The new null points will be located 11.1 cm on the normal bisector.