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Physics Question on Magnetic Field

An electron is revolving around a proton in a circular path of diameter 0.1 nm. It produces a magnetic field 14 Wb m−214\,Wb\,m^{-2} at the proton. The angular speed of the electron is

A

1.4×1016 rad s−11.4\times10^{16}\,rad\,s^{-1}

B

4.4×1016 rad s−14.4\times10^{16}\,rad\,s^{-1}

C

6.4×1016 rad s−16.4\times10^{16}\,rad\,s^{-1}

D

8.8×1016 rad s−18.8\times10^{16}\,rad\,s^{-1}

Answer

4.4×1016 rad s−14.4\times10^{16}\,rad\,s^{-1}

Explanation

Solution

B=μ04 π2 πRl=μ04 π×2 πet×IRB= \frac{\mu_0}{4 \, \pi } \frac{2 \, \pi}{R} l = \frac{\mu_0}{4 \, \pi} \times 2 \, \pi \frac{e}{t} \times \frac{I}{R} = μ04 π×2 π×e2 π R/v×1R=μ04 π×2 π×e v2 π R=μ04 πe ωR\frac{\mu_0}{4\, \pi} \times 2 \, \pi \times \frac{e}{2 \, \pi \,R/v} \times \frac{1}{R} = \frac{\mu_0}{4\, \pi} \times 2 \, \pi \times \frac{e\, v}{2 \, \pi \, R} = \frac{\mu_0}{4\, \pi } \frac{e \, \omega}{R}