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. $10\times {10}^{16}rad/sec$
• 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 $10\times {10}^{16}rad/sec$

Given: r=1$A^0$, B=16Wb/$m^2$

Solution: Formula of magnetic field at the centre of orbit of moving electron:

$B=\frac{{\mu }_{0}i}{2r}$

($t=\frac{2\pi }{\omega }$ )

Put values according to the given values:

$B=\frac{{\mu }_{0}q}{2rt}=\frac{{\mu }_0\left(ne\right)\times \omega }{2r\times 2\pi }$

$16=\frac{(4\pi \times {10}^{-7})\times 1\times 1.6\times {10}^{-19}\times \omega }{2\times {10}^{-10}\times 2\pi }$

Now we can get the angular velocity of electron:

$\omega =10\times {10}^{16}$

Hence the correct option is B