An electron moves around the nucleus in a hydrogen atom of radius 0.05 nm with a velocity of $2× 10^6 m s^{-1}$. The magnetic field produced at the centre of nucleus is :

12.8

The correct answer is Option (4) → 12.8

To calculate the magnetic field produced at a center of nucleus due to motion of an electron.

$B=\frac{μ_0I}{2r}$

where,

e - charge on electron

v - velocity of electron

r - radius of orbit ($0.05×10^{-9}m$)

Now,

$I=\frac{e}{T}=\frac{e×v}{2\pi r}$

where,

$I=\frac{(1.6×10^{-19C})(2×10^{6})}{2\pi(0.05×10^{-9})}$

$∴B=\frac{(4\pi×10^{-7})(1.6×10^{-19})(2×10^{6})}{4\pi(0.05×10^{-9})^2}$

$=12.8T$