Electromagnetic Waves JEE Main PYQs (2020–2026)

Solved Electromagnetic Waves PYQs

3 recent questions with answers. The full set of 125 is free to practise.

1. JEE Main 2026MediumEnergy, Intensity, and Radiation Pressure of EM Waves
   Q1

   Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
   
   Assertion (A) : The electromagnetic wave exerts pressure on the surface on which they are allowed to fall.
   
   Reason (R) : There is no mass associated with the electromagnetic waves.
   
   In the light of the above statements, choose the correct answer from the options given below :

   • ABoth (A) and (R) are true and (R) is the correct explanation of (A)
   • BBoth (A) and (R) are true but (R) is not the correct explanation of (A)
   • Cis true but (R) is false
   • Dis false but (R) is true

   Show answer & solution

   Correct answer: (B)

   Electromagnetic waves contain both energy (E) and momentum (p). And when an EM wave strikes a surface, it transfers its momentum to that surface.
   
   According to Newton’s Second Law, the rate of change of momentum is force, and force per unit area is pressure (radiation pressure).
   
   Therefore, assertion (A) is true.
   
   Electromagnetic waves consist of photons. Photons have zero rest mass. Though photons have energy and momentum, they do not have mass.
   
   Therefore, reason (R) is true.
   
   The reason EM waves exert pressure is because they carry momentum ( $p=\frac{E}{c}$ ), not because they don’t have mass. If waves had no momentum, they wouldn’t exert pressure regardless of their mass status. Thus, while reason is a true fact about EM waves, it does not explain why they exert pressure.
   
   So, both $(A)$ and $(R)$ are true, but $(R)$ is not the correct explanation of $(A)$ .
   
   Thus, the correct option is (B).
2. JEE Main 2026MediumDisplacement Current and Maxwell's Equations
   Q2

   A displacement current of 4.0 A can be set up in the space between two parallel plates of $6\mu F$ capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\times 10^{6}V/s$ . The value of $\alpha$ is $\_\_\_\_$ .

   • A0.58
   • B0.67
   • C0.82
   • D0.75

   Show answer & solution

   Correct answer: (B)

   According to Maxwell’s law, the displacement current in a capacitor is equal to the rate of change of charge on the plates $\left(I_d=\frac{dQ}{dt}\right)$ .
   
   The charge ( $Q$ ) on a capacitor of capacitance ( $C$ ), connected to potential difference (V) is:
   
   $$Q=CV$$
   
   On differentiating both sides with respect to time $(t)$ :
   
   $$\frac{dQ}{dt}=C\frac{dV}{dt}$$
   
   Since displacement current is $I_d=\frac{dQ}{dt}$ , we have:
   
   $$I_d=C\frac{dV}{dt}$$
   
   The displacement current is given as $I_d=4.0A$ and the capacitance of capacitor is $C=6\mu F=6\times 10^{-6}F$ : Putting the values,
   
   $$\frac{dV}{dt}=\frac{I_d}{C}$$
   
   $$\Rightarrow \frac{dV}{dt}=\frac{4.0}{6\times 10^{-6}}$$
   
   $$\Rightarrow \frac{dV}{dt}=\frac{2}{3}\times 10^6\approx 0.67\times 10^6\frac{V}{s}=\alpha \times 10^6\frac{V}{s}$$
   
   $$\Rightarrow \alpha =0.67$$
   
   Therefore, the correct option is (B).
3. JEE Main 2026EasyCharacteristics and Field Equations of Electromagnetic Waves
   Q3

   An electromagnetic wave travels in free space along the x-direction. At a particular point in space and time, $\overrightarrow{B}=2\times 10^{-7}\hat{j}$ T is associated with this wave. The value of corresponding electric field $\overrightarrow{E}$ at this point is _______ V/m.

   • A$60\text{mspace}6mu\hat{k}$
   • B$-60\text{mspace}6mu\hat{k}$
   • C$30\text{mspace}6mu\hat{k}$
   • D$-600\text{mspace}6mu\hat{k}$

   Show answer & solution

   Correct answer: (B)

   Given:
   
   Wave travels along $+x$ -direction
   
   Magnetic field is
   
   $$\overrightarrow{B}=2\times 10^{-7}\hat{j}\text{T}$$
   
   We know that for an electromagnetic wave,
   
   $$\overrightarrow{E}\times \overrightarrow{B}$$
   
   gives the direction of propagation.
   
   Since the wave is moving along $+\hat{i}$ and $\overrightarrow{B}$ is along $+\hat{j}$ , we need $\overrightarrow{E}$ such that
   
   $$\overrightarrow{E}\times \hat{j}=\hat{i}$$
   
   Now,
   
   $$(-\hat{k})\times \hat{j}=\hat{i}$$
   
   So, $\overrightarrow{E}$ must be along $-\hat{k}$ .
   
   Now use the relation in free space:
   
   $$E=cB$$
   
   where
   
   $$c=3\times 10^8\text{m/s}$$
   
   So,
   
   $$E=3\times 10^8\times 2\times 10^{-7}$$
   
   $$E=6\times 10^1=60\text{V/m}$$
   
   Therefore,
   
   $$\overrightarrow{E}=-60\hat{k}\text{V/m}$$
   
   So, the correct answer is:
   
   $$\boxed{-60\hat{k}\text{V/m}}$$
   
   Option B

Topic-wise question count

Where JEE Main has asked from Electromagnetic Waves, most frequent first.

• Characteristics and Field Equations of Electromagnetic Waves51
• Energy, Intensity, and Radiation Pressure of EM Waves34
• Propagation of Electromagnetic Waves in Matter16
• Electromagnetic Spectrum, Production, and Applications14
• Displacement Current and Maxwell's Equations10

Year-wise Electromagnetic Waves questions in JEE Main

Number of questions from this chapter in each year's paper(s).