Question

An electron is revolving around a proton in a circular orbit of diameter 1A. If it produces a magnetic field of $14wb/m^2$ at the proton, then its angular velocity will be about

• A. $8.75\times {10}^{16}$ rad/s
• B. ${10}^{10}$ rad/s
• C. $4\times {10}^{15}$ rad/s
• D. ${10}^{15}$ rad/s

Answer 1

The correct option is A $8.75\times {10}^{16}$ rad/s

$\left|B\right|=\frac{{\mu }_0\frac{e^{-}\omega }{2\pi }}{2r}$ where $e^{-}$=charge flowing ,$r$=charge on electron ,$\omega$=angular velocity(rad/sec)

where $\frac{e^{-}\omega }{2\pi }$ equivalent to current flowing in circular motion.

$14=\frac{4\pi \times {10}^{-7}\times \frac{1.6\times {10}^{-19}\times \omega }{2\pi }}{2(1\times {10}^{-10})}$

$14=1.6\times {10}^{-16}\omega$

$\omega =8.75\times {10}^{16}\frac{rad}{sec}$