Question

A bar magnet is $10cm$ long is kept with its north $(N)$ -pole pointing north. A neutral point is formed at a distance of $15cm$ 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

Answer: (B)

Length of magnet $=10cm=10\times 10^{-2}m$
$r=15\times 10^{-2}m$

$OP=\sqrt{225-25}=\sqrt{200}cm$
Since, at the neutral point, magnetic field due to the magnet is equal to $B_H$
$B_H=\frac{{\mu }_0}{4\pi }\cdot \frac{M}{{\left(O{P}^2+A{O}^2\right)}^{3/2}}$
$0.4\times 10^{-4}=10^{-7}\times \frac{M}{{\left(200\times 10^{-4}+25\times 10^{-4}\right)}^{3/2}}$
$\frac{0.4\times 10^{-4}}{10^{-7}}\times {\left(225\times 10^{-4}\right)}^{3/2}=M$
$0.4\times 10^3\times 10^{-6}(225{)}^{3/2}=M$
$M=1.35A-m$