Question

A bar magnet $20cm$ in length is placed with its south pole towards geographic north. The neutral points are situated al a distance of $40cm$ from centre of the magnet. If horizontal component of earth's field $3.2\times 10^{-5}T$, then pole strength of magnet is

• A. 5 Am
• B. 10 Am
• C. 45 Am
• D. 20 Am

Answer 1

Answer: (C)

Here, $2l=20cm$
$\Rightarrow l=10cm,d=40cm$
As, neutral point, $H=B=\frac{{\mu }_0}{4\pi }\frac{2Md}{{\left(d^2-l^2\right)}^2}$
$\Rightarrow 3.2\times 10^{-5}=\frac{10^{-7}\times 2M(0.4)}{15\times 15\times 10^{-4}}$
$\therefore M=\frac{3.2\times 15\times 15\times 10^{-4}\times 10^{-5}}{0.8\times 10^{-7}}=9$
$\therefore m=\frac{M}{2l}=\frac{9}{0.2}=45A-m$