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In the hydrogen atom, the electron is making $6.6 \times 10^{15} \mathrm{rps}$. If the radius of orbit is $0.53 \times 10^{-10} \mathrm{~m}$, then magnetic field produced at the centre of the orbit is
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The correct answer is:
$12.5 \mathrm{~T}$
Current, $i=q v$
$$
\begin{aligned}
B &=\frac{\mu_{0}^{i}}{2 r}=\frac{\mu_{0} q v}{2 r} \\
&=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19} \times 6.6 \times 10^{15}}{2 \times 0.53 \times 10^{-10}} \\
&=\frac{2 \pi \times 1.6 \times 6.6}{5.3}=12.513 \mathrm{~T}
\end{aligned}
$$
$$
\begin{aligned}
B &=\frac{\mu_{0}^{i}}{2 r}=\frac{\mu_{0} q v}{2 r} \\
&=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19} \times 6.6 \times 10^{15}}{2 \times 0.53 \times 10^{-10}} \\
&=\frac{2 \pi \times 1.6 \times 6.6}{5.3}=12.513 \mathrm{~T}
\end{aligned}
$$
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