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