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The magnitude of axial field due to a bar magnet at a distance of $1 \mathrm{~m}$, is found to be $5 \times 10^{-8} \mathrm{~T}$. The magnetic moment of the bar magnet is $\left(\mu_0=4 \pi \times 10^{-7}\right)$
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The correct answer is:
$0.25 \mathrm{Am}^2$
We have
$\begin{aligned} & \mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^3} \\ & \Rightarrow \quad 5 \times 10^{-8}=10^{-7} \times \frac{2 \times \mathrm{M}}{1^3} \\ & \Rightarrow \quad \mathrm{M}=0.25 \mathrm{Am}^2\end{aligned}$
$\begin{aligned} & \mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^3} \\ & \Rightarrow \quad 5 \times 10^{-8}=10^{-7} \times \frac{2 \times \mathrm{M}}{1^3} \\ & \Rightarrow \quad \mathrm{M}=0.25 \mathrm{Am}^2\end{aligned}$
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