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(B) The magnetic field at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.Since $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A

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$e\omega/r 	imes 10^{-7}$

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(B) The magnetic field at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.Since $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A$,we can write $B = \left(\frac{\mu_0}{4\pi}ight) \frac{2\pi I}{r}$.The current $I$ produced by an electron moving in a circular path is $I = \frac{e}{T}$,where $T$ is the time period.Since $T = \frac{2\pi}{\omega}$,the current is $I = \frac{e\omega}{2\pi}$.Substituting $I$ into the magnetic field formula:$B = (10^{-7}) 	imes \frac{2\pi (e\omega / 2\pi)}{r} = \frac{e\omega}{r} 	imes 10^{-7} \ Wb/m^2$.

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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