Question

A bar magnet $20\, cm$ in length is placed with its south pole towards geographic north. The neutral points are situated al a distance of $40\, cm$ 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, $2 l=20\, cm$
$\Rightarrow l=10\, cm ,\, d=40\, cm$
As, neutral point, $H=B=\frac{\mu_{0}}{4 \pi} \frac{2 Md }{\left( d ^{2}-l^{2}\right)^{2}}$
$\Rightarrow 3.2 \times 10^{-5}=\frac{10^{-7} \times 2 M(0.4)}{15 \times 15 \times 10^{-4}}$
$\therefore M=\frac{3.2 \times 15 \times 15 \times 10^{-4} \times 10^{-5}}{0.8 \times 10^{-7}}=9$
$\therefore m=\frac{M}{2 l}=\frac{9}{0.2}=45\, A - m$