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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2026 · 2 April · Morning Shift · Physics · Electromagnetic Waves

An electromagnetic wave travelling in x -direction is described by...

An electromagnetic wave travelling in xx-direction is described by field equation Ey=300sin⁡ω(t−xc)E_y = 300\sin\omega\left(t - \dfrac{x}{c}\right). If the electron is restricted to move in yy-direction only with speed of 1.5×1061.5 \times 10^6 m/s then ratio of maximum electric and magnetic forces acting on the electron is _______.

Options

  1. A200200
  2. B150150
  3. C400400
  4. D300300

Correct answer

A. 200200

Step-by-step solution

The maximum electric force acting on the electron is Fe=eE0F_e = eE_0.

The maximum magnetic force acting on the electron is Fm=evB0F_m = evB_0, where vv is the speed of the electron and B0B_0 is the maximum magnetic field.

The ratio of the maximum electric force to the maximum magnetic force is given by FeFm=eE0evB0=E0vB0\dfrac{F_e}{F_m} = \dfrac{eE_0}{evB_0} = \dfrac{E_0}{vB_0}.

For an electromagnetic wave, the ratio of the amplitudes of the electric and magnetic fields is E0B0=c\dfrac{E_0}{B_0} = c, where cc is the speed of light in vacuum.

Substituting this into the ratio of forces yields FeFm=cv\dfrac{F_e}{F_m} = \dfrac{c}{v}.

Given c=3×108c = 3 \times 10^8 m/s and v=1.5×106v = 1.5 \times 10^6 m/s, we get FeFm=3×1081.5×106=200\dfrac{F_e}{F_m} = \dfrac{3 \times 10^8}{1.5 \times 10^6} = 200.

Answer: 200200
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