Electromagnetic Waves JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Electromagnetic Waves, free to read — no sign-in needed. The full chapter has 4 questions; sign in to attempt the remaining 1 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   The electric field associated with an electromagnetic wave travelling in vacuum is given by $E_0 \sin(3y + 4z + \omega t)\,\hat{i}$, where $\omega$ is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are: [Given: speed of light in vacuum $c = 3 \times 10^8$ ms$^{-1}$.]
   1. A.The wave is travelling in $-\dfrac{1}{5}(3\hat{j} + 4\hat{k})$ direction.
   2. B.The magnitude of the wave vector is $0.5$ m$^{-1}$.
   3. C.The value of $\omega$ is $1.5 \times 10^9$ rad s$^{-1}$.
   4. D.The magnetic field associated with this wave is given by $\dfrac{E_0}{c}\sin(3y + 4z + \omega t)(4\hat{j} - 3\hat{k})$.
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   Answer: A,C

   The given electric field is $\vec{E} = E_0 \sin(3y + 4z + \omega t)\,\hat{i}$. Comparing the phase $\phi = 3y + 4z + \omega t$ with the standard wave equation phase $\vec{k} \cdot \vec{r} - \omega t$, we can rewrite it as $-(-3y - 4z - \omega t)$. Thus, the wave vector is $\vec{k} = -3\hat{j} - 4\hat{k}$. The direction of wave propagation is given by the unit vector $\hat{n}$: $\hat{n} = \dfrac{\vec{k}}{|\vec{k}|} = \dfrac{-3\hat{j} - 4\hat{k}}{\sqrt{(-3)^2 + (-4)^2}} = -\dfrac{1}{5}(3\hat{j} + 4\hat{k})$ The magnitude of the wave vector is $|\vec{k}| = 5$ m$^{-1}$. The angular frequency $\omega$ is: $\omega = c|\vec{k}| = (3 \times 10^8) \times 5 = 1.5 \times 10^9$ rad s$^{-1}$ The magnetic field $\vec{B}$ is given by: $\vec{B} = \dfrac{1}{c}(\hat{n} \times \vec{E}) = \dfrac{1}{c} \left[ -\dfrac{1}{5}(3\hat{j} + 4\hat{k}) \right] \times [E_0 \sin(3y + 4z + \omega t)\,\hat{i}]$ $\vec{B} = -\dfrac{E_0}{5c} \sin(3y + 4z + \omega t) [3(\hat{j} \times \hat{i}) + 4(\hat{k} \times \hat{i})]$ $\vec{B} = -\dfrac{E_0}{5c} \sin(3y + 4z + \omega t) (-3\hat{k} + 4\hat{j}) = \dfrac{E_0}{5c} \sin(3y + 4z + \omega t) (-4\hat{j} + 3\hat{k})$ Answer: The wave is travelling in $-\dfrac{1}{5}(3\hat{j} + 4\hat{k})$ direction.; The value of $\omega$ is $1.5 \times 10^9$ rad s$^{-1}$.
2. Q2JEE Advanced Adv 2025 (Paper 1)
   A cube of unit volume contains $35 \times 10^7$ photons of frequency $10^{15} \mathrm{~Hz}$. If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is $\alpha \times 10^{-9} \mathrm{~T}$. Taking permeability of free space $\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}$, Planck's constant $h=6 \times 10^{-34} \mathrm{Js}$ and $\pi=\frac{22}{7}$, the value of $\alpha$ is_______
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   Answer: 22.98

   Total energy in cube $=35 \times 10^7 \times \mathrm{hf}$ $\begin{aligned} & =35 \times 10^7 \times 6 \times 10^{-34} \times 10^{15} \\ & =2.1 \times 10^{-10} \mathrm{~J}\end{aligned}$ Total energy of EM waves $=\frac{B_0^2}{2 \mu_0} \times$ volume $\begin{aligned} & \mathrm{B}_0^2=\frac{2.1 \times 10^{-10} \times 8 \pi \times 10^{-7}}{1^3} \\ & \Rightarrow \mathrm{~B}_0=22.98 \times 10^{-9} \mathrm{~T}\end{aligned}$ Ans. 22.98
3. Q3JEE Advanced Adv 2008 (Paper 1)
   Which one of the following statement is WRONG in the context of X-rays generated from X-ray tube?
   1. A.Wavelength of characteristic X-rays decreases when the atomic number of the target increases
   2. B.Cut-off wavelength of the continuous X-rays depends on the atomic number of the target
   3. C.Intensity of the characteristics X-rays depends on the electrical power given to the X-ray tube
   4. D.Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube
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   Answer: (B)

   Cut-off wavelength depends on the applied voltage not on the atomic number of the target. Characteristic wavelengths depend on the atomic number of target.