In hydrogen atom an electron is making $6.6 \times 10^{15} \mathrm{rev} / \mathrm{s}$ around the nucleus of…
In hydrogen atom an electron is making $6.6 \times 10^{15} \mathrm{rev} / \mathrm{s}$ around the nucleus of radius $0.47 Å$. The magnetic field induction produced at the centre of the orbit is nearly
$0.14 w b m ~ m^{-2}$
$1.4 \mathrm{wb} \mathrm{m}^{-2}$
$14 w b m^{-2}$
$140 w b^{-2}$
Solution
In hydrogen atom, $\mathrm{f}=6.6 \times 10^{15} \mathrm{H}_{\mathrm{z}}, \mathrm{r}=0.47 \mathrm{~A}^{\circ}$
$\therefore \quad \mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}=\mathrm{ef}=1.6 \times 10^{-19} \times 6.6 \times 10^{15}=10.56 \times 10^4 \mathrm{~A}$
$\therefore \quad$ Magnetic field at the centre of the orbit is
$\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}=\frac{4 \pi \times 10^{-1} \times 10.56 \times 10^{-4}}{2 \pi \times 0.47 \times 10^{-10}}=14 \mathrm{wbm}^{-2}$