An electron moves with a speed of \( 6\times10^{7}\,\mathrm{m\,s^{-1}} \) at right angles to the magnetic field of a plane electromagnetic wave travelling in vacuum. At that instant, the ratio of the magnitude of the electric force to the magnitude of the magnetic force exerted on the electron by the wave is (Take \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5
The wave acts on the electron through both its fields. The electric force is \( eE \). The electron moves at right angles to \( \vec{B} \), so the magnetic force is \( evB \). In vacuum, the fields of the wave follow \( E=cB \).
The data are:
\( v=6\times10^{7}\,\mathrm{m\,s^{-1}} \)
\( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \)
The ratio of the two forces is:
\( \begin{aligned} \frac{F_E}{F_B} &= \frac{eE}{evB} \\ &= \frac{cB}{vB} \\ &= \frac{c}{v} \\ &= \frac{3\times10^{8}}{6\times10^{7}} \\ &= 5 \end{aligned} \)
Therefore, the electric force on the electron is 5 times the magnetic force. The magnetic force of the wave becomes comparable to its electric force only for charges moving close to \( c \).
NCERT (2026-27):
Nature of Electromagnetic Waves – “… It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as”
\( B_0=(E_0/c) \)\( \qquad (8.10) \)