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A bar magnet of length $6 \mathrm{~cm}$. is placed in the magnetic meridian with $N$ pole pointing towards the geographical north. Two neutral points, separated by a distance of $\mathrm{S}$ cm alle obtained on the equitorial axis of the magnet. If $B_H=1.2 \times 10^{-5}$ T. Then the pole strength of the magnet is
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The correct answer is:
$0.25 \mathrm{~A}^{-111^2}$
At neutral point
$$
\begin{array}{ll}
B & =B_H \\
\Rightarrow & \frac{\mu_0}{4 \pi} \times \frac{M}{\left(r^2+l^2\right)^{3 / 2}}=B_H \\
\Rightarrow & 10^{-7} \times \frac{m \times 2 l}{\left(r^2+l^2\right)^{3 / 2}}=1.2 \times 10^{-5} \\
\Rightarrow & 10^{-7} \times \frac{m \times 2 \times 3 \times 10^{-2}}{\left[\left(4 \times 10^{-2}\right)^2+\left(3 \times 10^{-2}\right)^2\right]^{3 / 2}}=1.2 \times 10^{-5}
\end{array}
$$
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{6 m \times 10^{-4}}{\left[\left(5 \times 10^{-2}\right)^2\right]^{3 / 2}} & =1.2 \\
\Rightarrow & & \frac{6 m \times 10^{-4}}{125 \times 10^{-6}} & =1.2 \\
\Rightarrow & & m & =\frac{1.2 \times 125 \times 10^{-2}}{6} \\
& & =0.25 \mathrm{~A}-\mathrm{m}^2
\end{array}
$$
$$
\begin{array}{ll}
B & =B_H \\
\Rightarrow & \frac{\mu_0}{4 \pi} \times \frac{M}{\left(r^2+l^2\right)^{3 / 2}}=B_H \\
\Rightarrow & 10^{-7} \times \frac{m \times 2 l}{\left(r^2+l^2\right)^{3 / 2}}=1.2 \times 10^{-5} \\
\Rightarrow & 10^{-7} \times \frac{m \times 2 \times 3 \times 10^{-2}}{\left[\left(4 \times 10^{-2}\right)^2+\left(3 \times 10^{-2}\right)^2\right]^{3 / 2}}=1.2 \times 10^{-5}
\end{array}
$$
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{6 m \times 10^{-4}}{\left[\left(5 \times 10^{-2}\right)^2\right]^{3 / 2}} & =1.2 \\
\Rightarrow & & \frac{6 m \times 10^{-4}}{125 \times 10^{-6}} & =1.2 \\
\Rightarrow & & m & =\frac{1.2 \times 125 \times 10^{-2}}{6} \\
& & =0.25 \mathrm{~A}-\mathrm{m}^2
\end{array}
$$
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