In an ionised sodium atom, an electron is moving in a circular path of radius $r$ with angular velocity $\omega $. The magnetic induction in $wb/m^2$ produced at the nucleus will be
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$B=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 \pi I}{r}$
Here $1=\frac{e}{T}=\frac{e \omega}{2 \pi} \therefore B=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 \pi e \omega / 2 \pi}{r}$
i. $e . \mathrm{B}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{e \omega}{r}=10^{-7}=\frac{e \omega}{r} \times 10^{-7} T$
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