A bar magnet is $10 \mathrm{~cm}$ long is kept with its north $(N)$-pole pointing north. A neutral point is formed at a distance of $15 \mathrm{~cm}$ 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
• Correct answer
  B

  6.75 A-m
• C

  27 A-m
• D

  1.35 A-m

Solution

$\begin{aligned} & \text { Length of magnet }=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}, \\ & \qquad r=15 \times 10^{-2} \mathrm{~m}\end{aligned}$
$O P=\sqrt{225-25}=\sqrt{200} \mathrm{~cm}$ Since, at the neutral point, magnetic field due to the magnet is equal to $B_H$ $\begin{gathered} B_H=\frac{\mu_0}{4 \pi} \cdot \frac{M}{\left(O P^2+A O^2\right)^{3 / 2}} \\ 0.4 \times 10^{-4}=10^{-7} \times \frac{M}{\left(200 \times 10^{-4}+25 \times 10^{-4}\right)^{3 / 2}} \\ \frac{0.4 \times 10^{-4}}{10^{-7}} \times\left(225 \times 10^{-4}\right)^{3 / 2}=M \\ 0.4 \times 10^3 \times 10^{-6}(225)^{3 / 2}=M \\ M=1.35 \mathrm{~A}-\mathrm{m} \end{gathered}$

Hence, the correct answer is option: 'B'.