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# What is the magnetic energy density (in terms of standard condition and r) at the center of a circulating electron in the hydrogen atom in first orbit.(radius of the orbit is r).

Last updated date: 15th Jul 2024
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Hint: We can try to apply the balanced condition that is to take a circular path centripetal force as well as electrostatic force is important.

According to the Bohr atom model, the negatively charged electron revolves around the positively charged nucleus in a circular orbit. The centripetal force on the electron is the electrostatic force of attraction on it due to the nucleus.
The electron of charge (-e) performs uniform circular motion around the nucleus of charge +Ze. Let r be the radius of the orbit of an electron and v be the orbital speed.
Let us consider the current due to motion of the electron is $I = \dfrac{e}{T}$…………. (1)
Where T is the period of revolution. We have $time = \dfrac{{dis\tan ce}}{{orbital{\text{ }}velocity}}$
$T = \dfrac{{2\pi r}}{{v}}$
$\therefore I = \dfrac{{ev}}{{2\pi r}}$ ……………….(1)
Let us consider the Magnetic field at the center of the loop ( it is like circular coil of one turn only)$B = \dfrac{{{\mu _0}I}}{{2r}}$
Substitute equation (1) in the above equation we get, $B = \dfrac{{{\mu _0}\left( {\dfrac{{ev}}{{2\pi r}}} \right)}}{{2r}} = \dfrac{{{\mu _0}ev}}{{4\pi {r^2}}}$…………… (2)
Therefore, the Magnetic energy density is$= \dfrac{{{B^2}}}{{2{\mu _0}}}({\text{standard formula}})$
Then substitute equation (2) in the above equation, we get,
Magnetic energy density $= \dfrac{1}{{2{\mu _0}}}\left[ {\dfrac{{{\mu _0}^2{v^2}{e^2}}}{{16{\pi ^2}{r^4}}}} \right]$
$= \dfrac{{{\mu _0}{v^2}{e^2}}}{{32{\pi ^2}{r^4}}}$
This is the magnetic energy density (in terms of standard condition and r) at the center of a circulating electron in the hydrogen atom in first orbit.

The magnetic dipole moment of a magnetic dipole is the product of the strength of its either pole and magnetic length. The magnetic dipole moment is a vector directed from south pole to north pole. The magnetic dipole moment has the S.I unit $A{m^2}$ .
According to Bohr atom model, the angular momentum $l = \dfrac{{nh}}{{2\pi }}$