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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).

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Last updated date: 15th Jul 2024
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Answer
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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.

Complete step-by-step answer:
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.

Additional Information:
Magnetic moment of a current loop is a vector perpendicular to the plane of the loop and the direction is given by right hand thumb rule.
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 $A{m^2}$ .
The ratio of magnetic moment of the electron to the orbital angular momentum of the electron is called gyromagnetic ratio.
According to Bohr atom model, the angular momentum $l = \dfrac{{nh}}{{2\pi }}$
Where n=1,2,3,4,…….. and h is Planck’s constant

Note: To produce the current through the inductor we need some work. The magnetic energy stored in the magnetic field will be equal to that work to produce the current through the inductor.