An electron in a hydrogen atom revolves around its nucleus with a...

The magnetic field due to the electron is given by

$B = \dfrac{{\mu}_{0}qvsin90 °}{4\pi{r}^{2}}$

The given data is

$v = 6 . 76 \times {10}^{6}m{s}^{- 1}r = 0 . 52 \times {10}^{- 10}m$

Thus, the magnetic field is

$B = \dfrac{{10}^{- 7} \times \left(1 . 6 \times \times {10}^{- 19}\right) \times \left(6 . 76 \times {10}^{6}\right)}{0 . 52 \times 0 . 52 \times {10}^{- 20}} = 40T$