Derive an expression for the magnetic field at the site of the nucleus in a hydrogen atom due to the circular motion of the electron Assume that the atom is in its ground state and the answer in terms of fundamental constants

Explanation

`i=(q)/(T)=(e)/(2pir//v)=(ev)/(2pi r)`
`B=(mu_(0i))/(2r)=(mu_(0)ev)/(4pi r^(2))`
`(mv^(2))/(r )=(1)/(4pi epsilon_(0)).(e.e_/(r^(2))`
`v=((e^(2))/(4pi epsilon_(0) mr))^(1)/(2)`
`mvr=(nh)/(2pi)`
`m((e^(2))/(4pi epsilon mr))^(1//2) r=(nh)/(2pi)`
`m^(2)(e^(2))/(4pi epsilon_(0)mr).r^(2)=(n^(2)h^(2))/(4pi^(2))`
`r=r_(n)=(n^(2)h^(2)epsilon_(0))/(pi me^(2))`
`v=v_(n)=sqrt((e^(2))/(4pi epsilon_(0) mr))=sqrt((e^(2))/(4pi epsilon_(0)m.(n^(2)h^(2)epsilon_(0))/(pi me^(2))))=(e^(2))/(2n epsilon_(0)h)`
`B=(mu_(0)ev_(n))/(4pi r_(n)^(2))=(mu_(0)e.e^(2))/(4 pi.2 n epsilon_(0)h.(n^(2)h^(4))/(pime^2))`
`=(mu_(0)e^(7)m^(2)pi)/(8n^(5) epsilon_(0)^(3)h^(5))`
` B =(mu_(0)e^(7)m^(2)pi)/(8 epsilon_(0)^(3)h^(5))`
For ground state, n=1
`B=(mu_(0)e^(7)m^(2)pi)/(8 epsilon_(0)^(3)h^(5))`