In a hydrogen atom, an electron moves in an orbit of radius \( 5 \times 10^{-11}\,\mathrm{m} \) with a speed of \( 2.2 \times 10^{6}\,\mathrm{m/s} \). The equivalent current is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.12 \times 10^{-3}\,\mathrm{A} \)
An electron going round an orbit carries its charge \( e \) past any point once every time period, so the equivalent current is the charge divided by the period.
\( \displaystyle T = \frac{2\pi r}{v} \;\Rightarrow\; I = \frac{e}{T} = \frac{e\,v}{2\pi r} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( r = 5\times10^{-11}\,\mathrm{m} \)
\( v = 2.2\times10^{6}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle I = \frac{1.6\times10^{-19} \times 2.2\times10^{6}}{2\pi \times 5\times10^{-11}} = \frac{3.52\times10^{-13}}{3.14\times10^{-10}} \)
\( \displaystyle I \approx 1.12\times10^{-3}\,\mathrm{A} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
The Magnetic Dipole Moment of a Revolving Electron – “… The electron of charge (–e) (e = + 1.6 × 10–19 C) performs uniform circular motion around a stationary heavy nucleus of charge +Ze. This constitutes a current I, where,
\( \displaystyle I = \frac{e}{T} \qquad (4.32) \)
and T is the time period of revolution. Let r be the orbital radius of the electron, and v the orbital speed. Then,
\( \displaystyle T = \frac{2\pi r}{v} \qquad (4.33) \)
Substituting in Eq. (4.32), we have \( I = ev/2\pi r \).”