Question

A bar magnet of magnetic moment $0.4A{m}^2$ is placed in the magnetic meridian with its north pole pointing north. A neutral point is obtained at a distance of $10cm$ from the centre of the magnet.If the length of the magnet is $20cm$, then the value of horizontal component of earth magnetic induction is:

• A. $\sqrt{2}\times {10}^{-5}T$
• B. $\frac{1}{\sqrt{2}}\times {10}^{-7}$
• C. $2\times {10}^{-7}T$
• D. $\frac{1}{2}\times {10}^{-7}T$

Answer 1

The correct option is A $\sqrt{2}\times {10}^{-5}T$

$m=0.4A{m}^2$
$l=20cm$
$d=10cm$
$m=lp$
$0.4=0.2p$
$P=2$
$|B_1|=|B_2|=B=\frac{{\mu }_{o}P}{4\pi d^2}$
$d^1=\sqrt{{10}^2+{10}^2}$
$B=\sqrt{B_1^2+B_2^2}$
$\sqrt{2}B$
$=\frac{\sqrt{2}4\pi \times {10}^{-7}\times 2}{4\pi \times \left(0.1\sqrt{2}R\right)}$
$=\sqrt{2}\times {10}^{-5}T$