Question

An electron is moving around a proton in an orbit of radius $1\mathring{A}$ and produces $16Wb/m^2$ of magnetic field at the centre, then find the angular velocity of electron:

• A. $20\pi \times {10}^{16}$ rad/sec
• B. ${10}^{17}$ rad/sec
• C. $\frac{5}{2\pi }\times {10}^{16}$ rad/sec
• D. $\frac{5}{4\pi }\times {10}^{16}$ rad/sec

Answer 1

The correct option is B ${10}^{17}$ rad/sec

$B=16Wb/m^2$

$R=\mathring{A}$

as $B=\frac{{\mu }_{0}I}{4\pi R^2}\oint dLB=\frac{{\mu }_{0}I}{4\pi R^2}\times 2\pi R=\frac{{\mu }_{0}I}{2R}$

as $I=\frac{e}{t}$ or $I=\frac{e\times \omega }{\pi }$ (as $\frac{2\pi }{\omega }=t$)

So, $16=\frac{4\pi \times {10}^{-7}\times (1.6\times {10}^{19})\times \frac{\omega }{2\pi }}{2\times {10}^{-10}}\Rightarrow \omega =\frac{16\times {10}^{-10}}{1.6\times {10}^{-26}}\Rightarrow \omega ={10}^{17}$rad/sec