In a hydrogen atom,find the magnetic field at | vedclass

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(C) The magnetic field $B$ at the center of the orbit due to an electron revolving in a circular path is given by $B = \frac{\mu_{0} I}{2r}$.The curre

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$13.95$

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(C) The magnetic field $B$ at the center of the orbit due to an electron revolving in a circular path is given by $B = \frac{\mu_{0} I}{2r}$.The current $I$ is given by $I = \frac{ev}{2\pi r}$,where $e$ is the charge of the electron,$v$ is the velocity,and $r$ is the radius.Substituting $I$ into the magnetic field formula: $B = \frac{\mu_{0} ev}{4\pi r^{2}}$.Given values: $\mu_{0} = 4\pi 	imes 10^{-7} \, T \cdot m/A$,$e = 1.6 	imes 10^{-19} \, C$,$v = 2.18 	imes 10^{6} \, m/s$,and $r = 5 	imes 10^{-11} \, m$.$B = \frac{(4\pi 	imes 10^{-7}) 	imes (1.6 	imes 10^{-19}) 	imes (2.18 	imes 10^{6})}{4\pi 	imes (5 	imes 10^{-11})^{2}}$$B = \frac{1.6 	imes 10^{-19} 	imes 2.18 	imes 10^{6} 	imes 10^{-7}}{25 	imes 10^{-22}}$$B = \frac{3.488 	imes 10^{-20}}{25 	imes 10^{-22}} = \frac{348.8}{25} = 13.952 \, T$.Thus,the magnetic field is approximately $13.95 \, T$.

In a hydrogen atom,find the magnetic field at the center in the ground state. Given that Bohr's radius is $r_{0} = 5 \times 10^{-11} \, m$ (in $T$).

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