Electromagnetic Waves: JEE Main 2026 Previous Year Questions
Solved JEE Main 2026 previous year questions for Electromagnetic Waves with clear step-by-step KaTeX solutions covering E-B relations, intensity, displacement current, and the EM spectrum.
Practice the most recent JEE Main 2026 Electromagnetic Waves questions with fully worked, step-by-step solutions.
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Solution
For an isotropic point source, the emitted power $P$ spreads uniformly over a sphere, so the intensity depends only on the distance $r$ from the source:
$$I = \frac{P}{4\pi r^2}$$Both the original and the new detector positions lie on the same spherical surface, meaning they are at the same distance $r = L$ from the source. The angular separation of $45^\circ$ only moves the detector along the sphere and does not change $r$.
$$I_{\text{new}} = \frac{P}{4\pi L^2} = I_0$$Answer: $I_0$
Solution
The peak electric field is $E_0 = 300\ \text{V/m}$, and the peak magnetic field of the wave is:
$$B_0 = \frac{E_0}{c}$$Maximum electric force on the electron (charge $q$):
$$F_E = qE_0$$Maximum magnetic force on the electron moving with speed $v$:
$$F_B = qvB_0 = qv\cdot\frac{E_0}{c}$$Taking the ratio, $E_0$ and $q$ cancel:
$$\frac{F_E}{F_B} = \frac{qE_0}{qv\,E_0/c} = \frac{c}{v} = \frac{3\times10^8}{1.5\times10^6} = 200$$Answer: 200
Solution
The phase $\left(2\pi vt - \dfrac{2\pi x}{\lambda}\right)$ shows the wave propagates along $+x$, so the propagation direction is $\hat{i}$.
For an EM wave, $\vec{E}$, $\vec{B}$ and the propagation direction $\hat{n}$ satisfy $\hat{E}\times\hat{B} = \hat{n}$. With $\vec{B}\parallel\hat{j}$ and $\hat{n}=\hat{i}$, we need $\hat{E}$ such that:
$$\hat{E}\times\hat{j} = \hat{i}$$Testing $\hat{E} = -\hat{k}$: $\;(-\hat{k})\times\hat{j} = -(\hat{k}\times\hat{j}) = -(-\hat{i}) = \hat{i}$ ✓
So the electric field points along $-\hat{k}$. Its magnitude uses $E_0 = cB_0$, and since $c = v\lambda$:
$$\vec{E} = -v\lambda B_0 \sin\left(2\pi vt - \frac{2\pi x}{\lambda}\right)\hat{k}$$Answer: $\vec{E} = -v\lambda B_0 \sin\left(2\pi vt - \dfrac{2\pi x}{\lambda}\right)\hat{k}$
Solution
The magnitude of the electric field is:
$$E = cB = (3\times10^8)(2\times10^{-7}) = 60\ \text{V/m}$$The propagation is along $+x$ ($\hat{i}$), and $\vec{B}\parallel\hat{j}$. Using $\hat{E}\times\hat{B} = \hat{n}$:
$$\hat{E}\times\hat{j} = \hat{i} \implies \hat{E} = -\hat{k}$$since $(-\hat{k})\times\hat{j} = \hat{i}$.
$$\vec{E} = -60\,\hat{k}\ \text{V/m}$$Answer: $-60\,\hat{k}$
Solution
For a plane EM wave, the magnetic field is obtained from $\vec{B} = \dfrac{\hat{k}\times\vec{E}}{c}$, where $\hat{k}$ is the unit propagation vector.
Since $\hat{k} = \dfrac{\vec{k}}{k}$ and $c = \dfrac{\omega}{k}$:
$$\vec{B} = \frac{1}{c}\left(\frac{\vec{k}}{k}\right)\times\vec{E} = \frac{k}{\omega}\cdot\frac{1}{k}\,(\vec{k}\times\vec{E}) = \frac{\vec{k}\times\vec{E}}{\omega}$$This correctly gives $\vec{B}$ perpendicular to both $\vec{k}$ and $\vec{E}$, with $\hat{E}\times\hat{B}$ pointing along $\vec{k}$.
Answer: $\dfrac{\vec{k}\times\vec{E}}{\omega}$
Solution
The displacement current equals the conduction current charging the capacitor:
$$I_d = C\,\frac{dV}{dt}$$Solving for the rate of change of potential difference:
$$\frac{dV}{dt} = \frac{I_d}{C} = \frac{4.0}{6\times10^{-6}} = 0.667\times10^{6}\ \text{V/s}$$Comparing with $\alpha\times10^6$ V/s gives $\alpha \approx 0.67$.
Answer: $0.67$
Solution
Assertion (A): EM waves carry linear momentum. When they fall on a surface, momentum is transferred, producing radiation pressure. For a fully absorbing surface, $P_{\text{rad}} = \dfrac{I}{c}$. So (A) is true.
Reason (R): Although photons have zero rest mass, an EM wave still carries momentum $p = \dfrac{E}{c}$ and an associated effective (relativistic) mass $m = \dfrac{E}{c^2}$. It is precisely this momentum that causes the pressure. The blanket claim “there is no mass associated with EM waves” is misleading/incorrect and, moreover, does not explain why pressure arises. So (R) is false.
Therefore, (A) is true but (R) is false.
Answer: (A) is true but (R) is false
Solution
Energy per photon = total power ÷ number of photons per second:
$$E = \frac{P}{N} = \frac{15\times10^{3}}{2.5\times10^{22}} = 6\times10^{-19}\ \text{J}$$Wavelength from $E = \dfrac{hc}{\lambda}$:
$$\lambda = \frac{hc}{E} = \frac{(6.6\times10^{-34})(3\times10^{8})}{6\times10^{-19}} = 3.3\times10^{-7}\ \text{m} = 330\ \text{nm}$$The visible range is roughly $400$–$700$ nm. Since $330\ \text{nm} < 400\ \text{nm}$, this radiation lies in the ultraviolet region.
Answer: Ultraviolet