Electromagnetic Waves JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Electromagnetic Waves, free to read — no sign-in needed. The full chapter has 9 questions; sign in to attempt the remaining 4 in the exam simulator.

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 $\dfrac{E_0}{c}\sin(3y + 4z + \omega t)(4\hat{j} - 3\hat{k})$.
   Show answer & solution

   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_______
   Show answer & solution

   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 2024 (Paper 2)
   A metal target with atomic number $Z=46$ is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio $r$ of the wavelengths of the $K_\alpha$-line and the cut-off is found to be $r=2$. If the same electron beam bombards another metal target with $Z=41$, the value of $r$ will be
   1. A.2.53
   2. B.1.27
   3. C.2.24
   4. D.1.58
   Show answer & solution

   Answer: (A)

   $\begin{aligned} & \frac{1}{\lambda_\alpha}=\frac{3}{4} R(Z-1)^2 p \\ & \lambda_{\text {cut }}=\frac{h c}{e V} \\ & \Rightarrow \text { Ratio } \propto \frac{1}{(Z-1)^2} \text { for same beam } \\ & \frac{Z}{x}=\frac{40^2}{45^2} \\ & \Rightarrow x=\frac{45^2}{40^2} .2 \approx 2.53 \end{aligned}$
4. Q4JEE Advanced Adv 2020 (Paper 2)
   In an $X$-ray tube, electrons emitted from a filament (cathode) carrying current $I$ hit a target (anode) at a distance $d$ from the cathode. The target is kept at a potential $V$ higher than the cathode resulting in emission of continuous and characteristic X-rays. If the filament current $I$ is decreased to $\dfrac{I}{2},$ the potential difference $V$ is increased to $2V,$ and the separation distance $d$ is reduced to $\dfrac{d}{2},$ then
   1. A.the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same
   2. B.the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain the same
   3. C.the cut-off wavelength will reduce to half, and the intensities of all the X-rays will decrease
   4. D.the cut-off wavelength will become two times larger, and the intensity of all the X-rays will decrease
   Show answer & solution

   Answer: A,C

   Cut off wavelength ${\lambda }_{C}\propto \dfrac{1}{V}$ So, cut-off wavelength becomes half. Characteristic x-ray depends on target atomic number, so it remains same. On decreasing filament current number of electron decrease, so intensity of $X$ -ray decreases.
5. Q5JEE Advanced Adv 2015 (Paper 2)
   A fission reaction is given by $U92236\rightarrow Xe54140+Sr3894+x+y$ , where $x$ and $y$ are two particles. Considering $U92236$ to be at rest, the kinetic energies of the products are denoted by ${K}_{Xe},{K}_{Sr},{K}_{x}(2MeV)$ and ${K}_{y}\left(2MeV\right)$ , respectively. Let the binding energies per nucleon of $U92236,Xe54140$ and $Sr3894$ be 7.5 MeV, 8.5 MeV, and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is (are)
   1. A.$x=n,y=n,{K}_{Sr}=129MeV,{K}_{Xe}=86MeV$
   2. B.$x=p,y={e}^{-},{K}_{Sr}=129MeV,{K}_{Xe}=86MeV$
   3. C.$x=p,y=n,{K}_{Sr}=129MeV,{K}_{Xe}=86MeV$
   4. D.$x=n,y=n,{K}_{Sr}=86MeV,{K}_{Xe}=129MeV$
   Show answer & solution

   Answer: (A)

   $\begin{matrix}U\rightarrow Xe+Sr+ & x & + \\ & 2 & \end{matrix}\begin{matrix}y \\ 2\end{matrix}$ $Q=4+{K}_{Xe}+{K}_{Sr}$ ...(i) $-Q={E}_{B}=236\times 7.5-140\times 8.5-94\times 8.5$ $∴Q=219$ ...(ii) $∴{K}_{Xe}+{K}_{Sr}=215MeV$ Since, both $x$ & $y$ have same KE $∴$ both particles should have same mass & lighter body will have higher KE.