In an ionised sodium atom, an electron is mov | vedclass

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Question

$B=\left(\frac{\mu_{0}}{4 \pi}ight) \frac{2 \pi I}{r}$

Here $1=\frac{e}{T}=\frac{e \omega}{2 \pi} 	herefore B=\left(\frac{\mu_{0}}{4 \pi}ight)

Vedclass · Accepted Answer

$e\omega/r 	imes10^{-7}$

Vedclass · Answer

$B=\left(\frac{\mu_{0}}{4 \pi}ight) \frac{2 \pi I}{r}$

Here $1=\frac{e}{T}=\frac{e \omega}{2 \pi} 	herefore B=\left(\frac{\mu_{0}}{4 \pi}ight) \frac{2 \pi e \omega / 2 \pi}{r}$

i. $e . \mathrm{B}=\left(\frac{\mu_{0}}{4 \pi}ight) \frac{e \omega}{r}=10^{-7}=\frac{e \omega}{r} 	imes 10^{-7} T$

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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