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In hydrogen a atom, an electron is revolving in the orbit of radius $0.53 Å$ with $6.6 \times 10^{15}$ radiations/s. Magnetic field produced at the centre of the orbit is
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The correct answer is:
$12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
The magnetic field
$$
\begin{array}{l}
\mathrm{B}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi(\mathrm{qv})}{\mathrm{r}} \\
=10^{-7} \times \frac{2 \times 3.14 \times\left(1.6 \times 10^{-19} \times 1.6 \times 10^{15}\right)}{0.53 \times 10^{-10}} \\
=12.5 \mathrm{~Wb} / \mathrm{m}^{3}
\end{array}
$$
$$
\begin{array}{l}
\mathrm{B}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi(\mathrm{qv})}{\mathrm{r}} \\
=10^{-7} \times \frac{2 \times 3.14 \times\left(1.6 \times 10^{-19} \times 1.6 \times 10^{15}\right)}{0.53 \times 10^{-10}} \\
=12.5 \mathrm{~Wb} / \mathrm{m}^{3}
\end{array}
$$
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