Electromagnetic Waves MCQ For NEET Physics with Answers

Why Practice NEET Electromagnetic Waves MCQs?

Practising NEET Electromagnetic Waves MCQs regularly helps build strong conceptual clarity and improve problem-solving speed for the exam. Strengthens understanding of wave concepts and formulas Improves accuracy in numerical and assertion-based questions Helps identify important NEET physics questions and frequently asked topics Enhances time management during the exam Reduces the chances of negative marking through better practice.

Download Electromagnetic Waves MCQs PDF for NEET 2026

Get access to a free downloadable PDF of Electromagnetic Waves MCQs with answers and detailed solutions for effective revision. This PDF is ideal for last-minute practice, quick revision of important concepts, and solving questions on the go. Students can use it to revise key formulas, strengthen weak areas, and boost their NEET 2026 Physics preparation. Questions Download Link NEET Physics 2026 Electromagnetic Waves MCQs with Solutions Download Here

Electromagnetic Waves MCQs for NEET 2026: Practice Questions with Answers

Improve your preparation with these carefully selected Electromagnetic Waves MCQs made as per the latest NEET exam pattern. These questions cover important concepts such as wave propagation, electromagnetic spectrum, radiation pressure, and field relationships. Practising them regularly will help you improve accuracy, speed, and confidence in solving both conceptual and numerical problems.

Q1. A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:

(a) zero between the plates and non-zero outside

(b) zero at all paces

(c) constant between the plates and zero outside the plates

(d) non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

Correct Answer: (d) non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

Explanation:

$\begin{aligned} & \mathrm{I}_{\mathrm{d}}=\epsilon_0 \frac{\mathrm{~d} \phi_e}{\mathrm{dt}}=\epsilon_0 \frac{\mathrm{~d}}{\mathrm{dt}}\left(\mathrm{EA} \cos 0^{\circ}\right) \\ & \mathrm{I}_{\mathrm{d}}=\epsilon_0 \mathrm{~A} \frac{\mathrm{dE}}{\mathrm{dt}} \\ & \because \mathrm{E}=\frac{\sigma}{\epsilon_0} ; \mathrm{I}_{\mathrm{d}}=\epsilon_0 \mathrm{~A} \frac{\mathrm{~d}}{\mathrm{dt}}\left(\frac{\sigma}{\epsilon_0}\right) \\ & \Rightarrow \mathrm{I}_{\mathrm{d}}=\mathrm{A}\left(\frac{\mathrm{d} \sigma}{\mathrm{dt}}\right)=\text { constant }\end{aligned}$

Also, $B$ due to $L_d$ is maximum at the surface of an imaginary cylinder.

Q2. A capacitor of capacitance $C$, is connected across an AC source of voltage $V$, given by $ V=V_0 \sin \omega t $ The displacement current between the plates of the capacitor, would then be given by

(a) $I_d=V_0 \omega C \cos \omega t$

(b) $I_d=\frac{V_0}{\omega C} \cos \omega t$

(c) $I_d=\frac{V_0}{\omega C} \sin \omega t$

(d) $I_d=V_0 \omega C \sin \omega t$

Correct answer: (a) $I_d=V_0 \omega C \cos \omega t$

Explanation: Given, AC source voltage, $ V=V_0 \sin \omega t $

We know that, $ 0=C V $ Here, $Q$ is the charge on the capacitor,

$C$ is the capacitance of the capacitor,

$V$ is the AC source voltage.

On differentiate Eq. (ii) w.r.t. time, we get

$ \begin{aligned} \frac{d Q}{d t} & =\frac{d(C V)}{d t} \\ \Rightarrow \frac{d Q}{d t} & =\frac{C d\left(V_0 \sin \omega t\right)}{d t} \quad \text { [from Eq.(i)] } \\ \Rightarrow \frac{d Q}{d t} & =C \omega V_0 \cos \omega t \end{aligned} $

As we know, the displacement current,

$ \begin{aligned} \quad I_d & =\frac{d Q}{d t} \\ \Rightarrow \quad I_d & =V_0 \omega C \cos \omega t \end{aligned} $

Q3. For a plane electromagnetic wave propagating in $x$-direction, which one of the following combination gives the correct possible directions for electric field ( $E$ ) and magnetic field ( $B$ ) respectively?

(a) $\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{j}}+\hat{\mathbf{k}}$

(b) $-\hat{\mathbf{j}}+\hat{\mathbf{k}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$

(c) $\hat{\mathbf{j}}+\hat{\mathbf{k}},-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

(d) $-\hat{\mathbf{j}}+\hat{\mathbf{k}},-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

Correct answer:

(b) $-\hat{\mathbf{j}}+\hat{\mathbf{k}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$

Explanation: We know that, in electromagnetic wave, the electric field ( $E$ ) and magnetic field (B) are perpendicular to each other,

$ \mathbf{E} \cdot \mathbf{B}=0 $ Consider the option (a);

$ (\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{j}}+\hat{\mathbf{k}})=1+1=2 \neq 0 $

So, it is incorrect option. Consider the option (b); $ (-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(-\hat{\mathbf{j}}-\hat{\mathbf{k}})=1-1=0 $

Hence, it satisfies the condition $\mathbf{E} \cdot \mathbf{B}=0$ Similarly, options (c) and (d) are incorrect.

Q4. The magnetic field in a plane electromagnetic wave is given by $ B_y=2 \times 10^{-7} \sin \left(\pi \times 10^3 x+3 \pi \times 10^{11} t\right) \mathrm{T} $ Calculate the wavelength. (a) $\pi \times 10^3 \mathrm{~m}$ (b) $2 \times 10^{-3} \mathrm{~m}$ (c) $2 \times 10^3 \mathrm{~m}$ (d) $\pi \times 10^{-3} \mathrm{~m}$

Correct answer: (b) $2 \times 10^{-3} \mathrm{~m}$

Explanation: Magnetic field in plane electromagnetic wave is given as:

$ B_y=2 \times 10^{-7} \sin \left(\pi \times 10^3 x+3 \pi \times 10^{11} t\right) \mathbf{T} $

Comparing with, $B_y=B_0 \sin (k x-\omega t)$, we get

$ \begin{aligned} & k=\pi \times 10^3 \Rightarrow \frac{2 \pi}{\lambda}=\pi \times 10^3 \\ \Rightarrow \quad \lambda & =\frac{2}{10^3}=2 \times 10^{-3} \mathrm{~m} \end{aligned} $

Q5. The ratio of contributions made by the electric field and magnetic field components, to the intensity of an electromagnetic wave is (where, $c=$ speed of electromagnetic waves)

(a) $1: 1$

(b) $1: c$

(c) $1: c^2$

(d) $c: 1$

Correct answer: (d) $c: 1$

Explanation: We know that, $ \frac{E_0}{B_0}=c $

where, $E_0$ and $B_0$ are the peak values of electric field and magnetic field, respectively.

$ \therefore \quad E_0: B_0=c: 1 $

Q6. Light with an average flux of 20 W/cm² falls on a non-reflecting surface at normal incidence having surface area $20 \mathrm{~cm}^2$. The energy received by the surface during time span of 1 min is

(a) $12 \times 10^3 \mathrm{~J}$

(b) $24 \times 10^3 \mathrm{~J}$

(c) $48 \times 10^3 \mathrm{~J}$

(d) $10 \times 10^3 \mathrm{~J}$

Correct answer: (b) $24 \times 10^3 \mathrm{~J}$

Explanation: Given, average flux $=20 \mathrm{~W} / \mathrm{cm}^2$ Surface area $=20 \mathrm{~cm}^2$ $ \text { Time }=1 \mathrm{~min}=60 \mathrm{~s} $ For non-reflecting surface, energy received $=$ average flux × surface area × time $ =20 \times 20 \times 60=24 \times 10^3 \mathrm{~J} $

How These MCQs Help in the NEET Electromagnetic Waves Chapter?

To score well in NEET Physics, practising Electromagnetic Waves MCQs is essential for building both conceptual clarity and application skills. This chapter is often tested through direct NCERT-based and formula-driven questions, so regular practice helps you stay exam-ready for NEET 2026.

• Familiarises you with the latest NEET exam pattern and commonly asked question types
• Strengthens core concepts like electromagnetic spectrum, displacement current, and wave properties
• Improves accuracy and reduces silly mistakes through consistent practice
• Increases time management to solve questions efficiently in the exam
• Develops strong problem-solving skills for both numerical and conceptual questions
• Helps identify weak areas and clear doubts with the help of detailed solutions

NEET 2026 Preparation Tips for Electromagnetic Waves

A smart and focused strategy is essential to score well in the Electromagnetic Waves chapter of NEET Physics. This topic is largely concept- and formula-based, so regular revision and practice can significantly improve performance. By following the right exam-orientated approach, you can boost accuracy, avoid common mistakes, and build confidence for NEET 2026.

• Revise key NEET physics formulas regularly, especially those related to wave speed, energy, and electromagnetic spectrum order
• Practice time-bound mock tests to improve speed, accuracy, and exam temperament
• Focus on important concepts like polarisation, displacement current, and phase relationships
• Use concise formula sheets for quick and effective last-minute revision
• Analyse incorrect answers carefully to understand mistakes and avoid repeating them.