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  • An electron of charge e moves in a circular orbit of radius r around a nucleus. The magnetic field due to orbital motion of the electron at the site of the nucleus in B. The angular velocity <

An electron of charge e moves in a circular orbit of radius r around a nucleus. The magnetic field due to orbital motion of the electron at the site of the nucleus in B. The angular velocity \mathrm{\omega} of the electron is

Option: 1

\mathrm{\omega=\frac{2 \mu_0 e B}{4 \pi r} }


Option: 2

\mathrm{\omega=\frac{\mu_0 e B}{\pi r} }


Option: 3

\mathrm{\omega=\frac{4 \pi r B}{\mu_0 e} }


Option: 4

\mathrm{\omega=\frac{2 \pi r B}{\mu_0 e}}


Answers (1)

best_answer

An electron moving in a circular orbit is equivalent to a current carrying loop. As explained above, the current is

\mathrm{I=\nu e=\frac{e}{T}}

where T is the time period of the motion of the electron around the nucleus. If v is the speed of the electron,

\mathrm{T =\frac{2 \pi r}{v} }
\mathrm{\therefore \quad I =\frac{e v}{2 \pi r}=\frac{e \omega}{2 \pi} \quad(\because v=r \omega) }

Now, the magnetic field at the centre of the loop is

\mathrm{B =\frac{\mu_0 I}{2 r}=\frac{\mu_0 e \omega}{4 \pi r} \\ }
or    \mathrm{\quad \omega =\frac{4 \pi r B}{\mu_0 e}}

Hence the correct choice is (c).

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mansi

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