Question

A bar magnet is 10 cm long is kept with its north (N)-pole pointing north. A neutral point is formed at a distance of 15 cm from each pole:Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet is

• A. 9 A-m
• B. 6.75 A-m
• C. 27 A-m
• D. 1.35 A-m

Answer 1

The correct option is D 1.35 A-m

Length of magnet = 10 cm = 10 $\times$ 10${}^{-2}$ m,
r = 15 $\times$ 10${}^{-2}$ m
$op=\overline{225-25}=\overline{200}cm$
Since , at the neutral point , magnetic field due to the magnet is equal to$B_H$
$B_H=\frac{{\mu }_O}{4\pi }\cdot \frac{M}{(O{P}^2+A{O}^2{)}^{3/2}}$
$0.4\times {10}^{-4}={10}^{-7}\times \frac{M}{(200\times {10}^{-4}+25\times {10}^{-4}{)}^{3/2}}$
$\frac{0.4\times {10}^{-4}}{{10}^{-7}}\times (225\times {10}^{-4}{)}^{3/2}=M$
$0.4\times {10}^{-6}{225}^{3/2}=M$
$M=1.35A-m$