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.

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