Question

If magnetic field strength at the center of a hydrogen atom caused by an electron moving along the first Bohr orbit is $\frac{x}{2}\text{T}$. The value of $x$ nearly is (integer only)

Answer 1

The equivalent current due to circular motion of the electron is,

$I=\frac{e\omega }{2\pi }$

The magnetic field at the centre of the loop will be,

$B=\frac{{\mu }_{0}I}{2R}$

$B=\frac{{\mu }_{0}I}{2R}=\frac{{\mu }_0}{2}\times \frac{e\omega }{2\pi R}=\frac{{\mu }_0}{4\pi }\times \frac{ev}{R^2}$

Where, $R=\text{Radius of the first Bohr orbit}v=\text{Velocity of the electron in first Bohr orbit}$

$\therefore B={10}^{-7}\times 1.6\times {10}^{-19}\times \frac{2.18\times {10}^6}{(0.53\times {10}^{-10}{)}^2}$

$\therefore B\approx 12.5\text{T}=\frac{25}{2}\text{T}$

$\therefore n=25$