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

  $9 \mathrm{~A}-\mathrm{m}$
• B

  $6.75 \mathrm{~A}-\mathrm{m}$
• C

  $27 \mathrm{~A}-\mathrm{m}$
• Correct answer
  D

  $1.35 \mathrm{~A}-\mathrm{m}$

Solution

Here, length ofmagnet $=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}$, $r=15 \times 10^{-2} \mathrm{~m}$
Since, at the neutral point, magnetic field due to the magnet is equal to $B_{H}$, $$ \begin{array}{l} B_{H}=\frac{\mu_{0}}{4 \pi} \cdot \frac{M}{\left(O P^{2}+A O^{2}\right)^{3 / 2}} \\ 0.4 \times 10^{-4}=10^{-7} \times \frac{M}{\left(200 \times 10^{-4}+25 \times 10^{-4}\right)^{3 / 2}} \\ \Rightarrow \frac{0.4 \times 10^{-4}}{10^{-7}} \times\left(225 \times 10^{-4}\right)^{3 / 2}=M \\ \Rightarrow 0.4 \times 10^{3} \times 10^{-6}(225)^{3 / 2}=M \\ \Rightarrow M=1.35 \mathrm{~A}-\mathrm{m} \end{array} $$

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