Physics · Volume 1 · Chapter 5 · Samacheer Kalvi Grade 12

Samacheer Class 12 Physics - Electromagnetic Waves

Free Content
37 Book Back Q&A
Verified Answers

Complete book back questions and answers for Class 12 Physics Chapter 5 - Electromagnetic Waves. Use the MCQ cards and expandable solution cards for board exam revision.

Share on WhatsApp

Samacheer Kalvi Class 12 Physics Electromagnetic Waves book back solutions with concise explanations and a verified validation footer on every answer.

What's on this page
Multiple Choice Questions 15 Short Answer Questions 8 Long Answer Questions 9 Numerical Problems 5
Your Progress - Chapter 5: Electromagnetic Waves0% complete
Multiple Choice QuestionsMultiple Choice Questions15 questions
Q.1 The dimension of $1/(\mu_0\varepsilon_0)$ is
Answer: B: [L^{2} T^{-2}]

Since $1/(\mu_0\varepsilon_0)=c^2$, and speed squared has dimension $[L^2T^{-2}]$, the required dimension is $[L^{2}T^{-2}]$.

Q.2 If the amplitude of the magnetic field is $3\times10^{-6}\,$T, then amplitude of the electric field for an electromagnetic wave is
Answer: D: 900 V m^{-1}

In vacuum $E_0=cB_0$. With $c\approx3\times10^8\,$m/s and $B_0=3\times10^{-6}\,$T, $E_0\approx3\times10^8\times3\times10^{-6}=9\times10^2\,$V/m.

Q.3 Which of the following electromagnetic radiations is used for viewing objects through fog
Answer: D: infrared

Infrared radiation can penetrate some amounts of fog and is used in thermal imaging and night vision systems.

Q.4 Which of the following is false for electromagnetic waves
Answer: C: longitudinal

Electromagnetic waves are transverse (E and B perpendicular to direction of propagation), so the statement 'longitudinal' is false.

Q.5 Consider an oscillator which has a charged particle oscillating about its mean position with a frequency of 300 MHz. The wavelength of electromagnetic waves produced by this oscillator is
Answer: A: 1 m

Wavelength $\lambda=c/f$. With $f=300\,$MHz $=3\times10^8\,$Hz and $c\approx3\times10^8\,$m/s, $\lambda\approx1\,$m.

Q.6 The electric and magnetic fields, associated with an electromagnetic wave, propagating along negative X axis can be represented by
Answer: B: $\mathbf{E}=E_0\hat{k}$ and $\mathbf{B}=B_0\hat{j}$

For propagation along $-\hat{i}$, we need $\mathbf{E}\times\mathbf{B}=-\hat{i}$. With $\mathbf{E}=\hat{k}$ and $\mathbf{B}=\hat{j}$, $\hat{k}\times\hat{j}=-\hat{i}$, so option B is consistent.

Q.7 In an electromagnetic wave travelling in free space the rms value of the electric field is 3 V m^{-1}. The peak value of the magnetic field is
Answer: A: 1.414 × 10^{-8} T

E_{rms}=3 V/m so E_0=\sqrt{2}E_{rms}=4.243 V/m. Peak magnetic field $B_0=E_0/c\approx4.243/(3\times10^8)=1.414\times10^{-8}\,$T.

Q.8 An e.m. wave is propagating in a medium with a velocity $\mathbf{v}=\hat{i}$. The instantaneous oscillating electric field of this e.m. wave is along +y-axis, then the direction of oscillating magnetic field of the e.m. wave will be along:
Answer: C: +z direction

For propagation along +x, we need $\mathbf{E}\times\mathbf{B}=\hat{i}$. With $\mathbf{E}=\hat{j}$ and $\mathbf{B}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, so $\mathbf{B}$ is +z.

Q.9 If the magnetic monopole exists, then which of Maxwell’s equation is to be modified?
Answer: B: $\displaystyle\oint_s \mathbf{B}\cdot d\mathbf{A}=0$

Gauss's law for magnetism states $\oint_s\mathbf{B}\cdot d\mathbf{A}=0$, expressing no magnetic monopoles. If monopoles exist this equation would have a nonzero right-hand side (magnetic charge enclosed).

Q.10 Fraunhofer lines are an example of _______ spectrum.
Answer: B: line absorption

Fraunhofer lines are dark lines in the solar spectrum produced when cooler gases absorb specific wavelengths from the continuous spectrum, i.e. an absorption-line spectrum.

Q.11 Which of the following is an electromagnetic wave?
Answer: C: gamma-rays

Gamma rays are high-energy photons and hence electromagnetic. Alpha and beta rays are charged particle radiation.

Q.12 Which one of them is used to produce a propagating electromagnetic wave?
Answer: A: an accelerating charge

Changing (accelerating) charges produce time-varying fields that radiate electromagnetic waves. A uniformly moving or stationary charge does not radiate.

Q.13 If $E=E_0\sin[10^6 x -\omega t]$ be the electric field of a plane electromagnetic wave, the value of $\omega$ is
Answer: D: 3 × 10^{14} rad s^{-1}

Here wave number $k=10^6\,$m^{-1}. For EM waves $\omega=ck\approx(3\times10^8)(10^6)=3\times10^{14}\,$rad/s.

Q.14 Which of the following is NOT true for electromagnetic waves?
Answer: D: in vacuum, it travels with different speeds which depend on their frequency

In vacuum all electromagnetic waves travel at the same speed $c$, independent of frequency; hence option D is not true.

Q.15 The electric and magnetic fields of an electromagnetic wave are
Answer: A: in phase and perpendicular to each other

For a plane electromagnetic wave in free space, the electric and magnetic fields oscillate in phase and are perpendicular to each other and to the direction of propagation.

Short Answer QuestionsShort Answer Questions8 questions
Q.1What is displacement current?v
Solution

Maxwell added displacement current density $\mathbf{J}_d=\varepsilon_0\dfrac{\partial\mathbf{E}}{\partial t}$. In integral form Ampère–Maxwell law becomes $\displaystyle\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\text{enc}}+\mu_0\varepsilon_0\dfrac{d}{dt}\Phi_E$. This explains magnetic effects in regions with changing electric flux.

Answer:

Displacement current is the term $\varepsilon_0\dfrac{\partial \mathbf{E}}{\partial t}$ (or $\varepsilon_0\dfrac{d\Phi_E}{dt}$ in integral form) introduced by Maxwell to account for the rate of change of electric field in regions (e.g. between capacitor plates) where there is no conduction current; it contributes to the magnetic field exactly as a real current does.

Q.2What are electromagnetic waves?v
Solution

They consist of mutually perpendicular electric and magnetic fields in phase; examples include radio waves, light, X-rays, and gamma rays.

Answer:

Electromagnetic waves are self-sustaining transverse waves of oscillating electric and magnetic fields that propagate through space carrying energy and momentum; they satisfy Maxwell's equations and travel at speed $c=1/\sqrt{\mu_0\varepsilon_0}$ in vacuum.

Q.3Write down the integral form of modified Ampère’s circuital law.v
Solution

The extra term $\mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$ is the displacement current term which restores consistency with charge conservation.

Answer:

Ampère–Maxwell law (integral form): $\displaystyle\oint_{\ell}\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\rm enc}+\mu_0\varepsilon_0\dfrac{d}{dt}\Phi_E$, where $\Phi_E=\int_{S}\mathbf{E}\cdot d\mathbf{A}$ is electric flux through surface bounded by $\ell$.

Q.4Write notes on Gauss' law in magnetism.v
Solution

This is a fundamental Maxwell equation indicating net magnetic flux through any closed surface is zero. If magnetic monopoles existed the right-hand side would be magnetic charge enclosed.

Answer:

Gauss's law for magnetism states $\displaystyle\oint_S\mathbf{B}\cdot d\mathbf{A}=0$ for any closed surface $S$, implying there are no isolated magnetic charges (monopoles) and magnetic field lines are continuous closed loops.

Q.5Give two uses each of (i) IR radiation, (ii) Microwaves and (iii) UV radiation.v
Solution

IR detects heat and is used in sensors; microwaves heat dielectric materials and are used in radar links; UV kills bacteria and is used to sterilize water/surfaces.

Answer:

(i) IR: thermal imaging (night vision), remote control/communication. (ii) Microwaves: microwave ovens (heating), radar and long-distance communication. (iii) UV: sterilization/disinfection, photochemical processes and fluorescent lamps.

Q.6What are Fraunhofer lines? How are they useful in the identification of elements present in the Sun?v
Solution

Each element produces characteristic spectral lines; comparing solar absorption line wavelengths with laboratory emission/absorption lines reveals the Sun's elemental composition.

Answer:

Fraunhofer lines are dark absorption lines in the solar spectrum produced when specific wavelengths are absorbed by cooler gases in the Sun's atmosphere. By matching the wavelengths of these absorption lines to known atomic spectra, the elements present in the Sun's atmosphere can be identified.

Q.7Write notes on Ampere-Maxwell law.v
Solution

The displacement current term explains magnetic fields in regions without conduction current (e.g. between capacitor plates) and allows Maxwell's equations to predict electromagnetic waves.

Answer:

Ampère–Maxwell law generalizes Ampère's law by adding the displacement current term: $\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\rm enc}+\mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$. This term accounts for time-varying electric fields and ensures continuity of current and consistency with charge conservation.

Q.8Why are e.m. waves non-mechanical?v
Solution

Mechanical waves need particles of a medium to transfer energy; EM waves transmit energy via fields themselves, independent of matter.

Answer:

Electromagnetic waves are non-mechanical because they do not require a material medium to propagate; they are oscillations of electric and magnetic fields in space and can travel through vacuum.

Long Answer QuestionsLong Answer Questions9 questions
Q.1Write down Maxwell equations in integral form.v
Solution

These four integral relations describe how charges produce electric fields, absence of magnetic monopoles, how time-varying magnetic flux induces electric fields, and how currents and changing electric flux produce magnetic fields.

Answer:

Maxwell's equations (integral form):

- Gauss's law (electric): $\displaystyle\oint_S\mathbf{E}\cdot d\mathbf{A}=\dfrac{Q_{\rm enc}}{\varepsilon_0}$.

- Gauss's law (magnetic): $\displaystyle\oint_S\mathbf{B}\cdot d\mathbf{A}=0$.

- Faraday's law: $\displaystyle\oint_{\ell}\mathbf{E}\cdot d\mathbf{l}=-\dfrac{d}{dt}\Phi_B$, where $\Phi_B=\int_S\mathbf{B}\cdot d\mathbf{A}$.

- Ampère–Maxwell law: $\displaystyle\oint_{\ell}\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\rm enc}+\mu_0\varepsilon_0\dfrac{d}{dt}\Phi_E$, where $\Phi_E=\int_S\mathbf{E}\cdot d\mathbf{A}$.

Q.2Write short notes on (a) microwave (b) X-ray (c) radio waves (d) visible spectrumv
Solution

Each band has characteristic frequencies, sources, interactions with matter and practical applications; e.g., visible light is used in optics, X-rays for imaging, microwaves for heating and communication, radio waves for broadcasting and telecommunication.

Answer:

(a) Microwaves: Frequency range ~300 MHz to 300 GHz. Used in cooking (dielectric heating), radar, satellite communication, and microwave links. They penetrate clouds and are useful in remote sensing.

(b) X-rays: High-energy electromagnetic radiation (approx. 10^{16}–10^{19} Hz). Strongly penetrating, used in medical imaging (radiography), crystallography, and industrial inspection. Produced by accelerated electrons hitting targets or by inner-shell electronic transitions.

(c) Radio waves: Lowest-frequency part of EM spectrum (~kHz to GHz). Used for AM/FM broadcasting, TV, two-way communication, and navigation. Long wavelengths allow long-distance propagation via ionospheric reflection.

(d) Visible spectrum: Wavelengths ~400–700 nm corresponding to light perceptible by human eye. Contains colours from violet to red; important for vision, photosynthesis, and optical instruments.

Q.3Discuss the Hertz experiment.v
Solution

Hertz's experiments in 1887–88 confirmed Maxwell's theory by producing and detecting radio waves, measuring their wavelength and speed, and demonstrating wave properties (reflection, refraction, polarization). This established that light is an electromagnetic wave.

Answer:

Hertz demonstrated the existence of electromagnetic waves predicted by Maxwell. He used an oscillating spark-gap transmitter (an RLC oscillator producing oscillating charges) to generate radio-frequency EM waves and detected them with a receiving loop that produced sparks when in resonance. He showed that these waves underwent reflection, refraction, polarization, interference and had the same speed as light.

Key points: generation by accelerated charges, detection by induced oscillations in a resonant circuit, measurement of wavelength via standing wave patterns, verification that EM waves behave like light.

Q.4Explain the Maxwell’s modification of Ampere’s circuital law.v
Solution

The displacement current term acts like a real current in producing magnetic fields, ensures continuity equation $\nabla\cdot\mathbf{J}+\partial\rho/\partial t=0$ is satisfied, and leads to wave equations for fields allowing propagation of EM waves.

Answer:

Maxwell added the displacement current term to Ampère's law to resolve an inconsistency with charge conservation. Original Ampère's law $\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\rm enc}$ fails for cases like a charging capacitor: different surfaces bounded by same loop give different enclosed currents. Maxwell introduced displacement current density $\mathbf{J}_d=\varepsilon_0\partial\mathbf{E}/\partial t$. The modified law becomes $\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\rm enc}+\mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$, restoring consistency and symmetry between electric and magnetic fields.

Q.5Explain the importance of Maxwell’s correction.v
Solution

Without the displacement current term, Maxwell's theory would not predict self-propagating EM waves. The correction enabled predictions later experimentally confirmed, forming the basis of classical electrodynamics and optics.

Answer:

Maxwell's correction (displacement current) is important because:

- It resolves the inconsistency in Ampère's law for time-varying fields (e.g. between capacitor plates).

- It enforces charge conservation and the continuity equation.

- It symmetrizes Maxwell's equations between electric and magnetic fields.

- Crucially, it leads to the prediction of electromagnetic waves: combining curl equations yields wave equations with speed $c=1/\sqrt{\mu_0\varepsilon_0}$. Thus electromagnetic radiation and light are unified.

Q.6Write down the properties of electromagnetic waves.v
Solution

These properties follow from Maxwell's equations and the wave solutions for E and B fields.

Answer:

Properties of electromagnetic waves:

- Transverse: E and B are perpendicular to the direction of propagation and to each other.

- Mutually sustaining: time-varying E produces B and vice versa.

- Travel in vacuum at speed $c=1/\sqrt{\mu_0\varepsilon_0}$.

- Carry energy and momentum; energy density $u=\tfrac{1}{2}\varepsilon_0 E^2+\tfrac{1}{2\mu_0}B^2$ and Poynting vector $\mathbf{S}=\tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$ gives energy flux.

- In free space E and B are in phase and related by $E_0=cB_0$.

- Can be polarized, reflected, refracted, diffracted and interfered.

- Do not require a material medium (non-mechanical).

- Obey superposition principle.

Q.7Discuss the source of electromagnetic waves.v
Solution

A localized oscillating charge (e.g. dipole antenna) radiates power; the instantaneous power radiated by an accelerating charge is given by Larmor formula $P=\dfrac{q^2a^2}{6\pi\varepsilon_0 c^3}$ (non-relativistic).

Answer:

Electromagnetic waves are produced by time-varying currents and accelerating charges. An accelerating charge produces changing electric and magnetic fields that propagate outward as radiation. Common sources: oscillating charges in antennas (radio waves), transitions of electrons between energy levels in atoms (visible, UV, X-rays), and bremsstrahlung from decelerating charged particles (X-rays).

Q.8Explain the types of emission spectrum.v
Solution

Atomic emission lines are used for element identification; continuous spectra depend on temperature (Planck's law); molecular bands appear broadened due to many closely spaced transitions.

Answer:

Types of emission spectra:

- Continuous spectrum: emitted by hot dense objects (blackbody) showing a continuous spread of wavelengths.

- Line (discrete) emission spectrum: produced by atoms or ions when electrons transition from higher to lower energy levels; appears as discrete bright lines at characteristic wavelengths (e.g. atomic emission lines).

- Band emission spectrum: arises from molecular transitions where vibrational and rotational sub-levels produce groups (bands) of closely spaced lines; typical of molecules and some solids.

Each type depends on the nature of the emitter—atoms produce line spectra, molecules band spectra, and thermal sources continuous spectra.

Q.9Explain the types of absorption spectrum.v
Solution

By comparing absorption lines/bands with known atomic or molecular spectra, one can identify substances and infer conditions in stellar atmospheres or laboratory samples.

Answer:

Types of absorption spectra:

- Continuous absorption: a continuous range of wavelengths absorbed (e.g. broad attenuation by a continuum absorber).

- Line (discrete) absorption spectrum: produced when a cooler low-density gas absorbs specific wavelengths from a continuous background source, producing dark lines at characteristic wavelengths corresponding to electronic transitions (Fraunhofer lines in the solar spectrum).

- Band absorption spectrum: molecules absorb in bands due to vibrational/rotational transitions, producing groups of dark bands.

Absorption spectra provide information about the composition and physical conditions (temperature, pressure) of the absorbing medium.

Numerical ProblemsNumerical Problems5 questions
Q.1Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.v
Solution

When the conduction current in the connecting wire is 5 A charging the capacitor, the rate of change of electric flux between the plates produces an equal displacement current through any surface between plates. Thus $I_d=I_{\rm conduction}=5\,$A (by continuity of current).

Answer:

Displacement current $I_d=5\,$A

Q.2A transmitter consists of LC circuit with an inductance of $1\ \mu$H and a capacitance of $1\ \mu$F. What is the wavelength of the electromagnetic waves it emits?v
Solution

Oscillation angular frequency $\omega=1/\sqrt{LC}$. With $L=1\times10^{-6}\,$H and $C=1\times10^{-6}\,$F, $\omega=1/\sqrt{10^{-12}}=10^{6}\,$rad/s. Frequency $f=\omega/(2\pi)=10^6/(2\pi)\approx1.592\times10^{5}\,$Hz. Wavelength $\lambda=c/f\approx(3\times10^8)/(1.592\times10^{5})\approx1.884\times10^{3}\,$m.

Answer:

Approximately $1.884\times10^{3}\,$m

Q.3A pulse of light of duration $10^{-6}\,$s is absorbed completely by a small object initially at rest. If the power of the pulse is $60\times10^{-3}\,$W, calculate the final momentum of the object.v
Solution

Energy of pulse $E=Pt=(60\times10^{-3})(10^{-6})=6.0\times10^{-8}\,$J. Photons carry momentum $p=E/c$. Thus $p=(6.0\times10^{-8})/(3\times10^8)=2.0\times10^{-16}\,$kg m s^{-1}.

Answer:

Final momentum $p=2.0\times10^{-16}\,$kg m s^{-1}

Q.4Let an electromagnetic wave propagate along the x-direction, the magnetic field oscillates at a frequency of $10^{10}\,$Hz and has an amplitude of $10^{-5}\,$T, acting along the y-direction. Then, compute the wavelength of the wave. Also write down the expression for electric field in this case.v
Solution

Given frequency $f=10^{10}$ Hz, $\lambda=c/f=(3\times10^8)/(10^{10})=3\times10^{-2}\,$m. For a wave traveling in +x with $\mathbf{B}$ along +y, $\mathbf{E}$ must be along +z with $E_0=cB_0=3\times10^3\,$V/m. So $\mathbf{E}(x,t)=3\times10^3\sin\left(2\pi\times10^{10}t -\dfrac{2\pi}{\lambda}x\right)\hat{z}$ or equivalently $3\times10^3\sin(kx-\omega t)\,\hat{z}$ with $\omega=2\pi\times10^{10}$ and $k=2\pi/\lambda$ (sign convention depending on propagation phase).

Answer:

Wavelength $\lambda=3\times10^{-2}\,$m. Electric field: $\mathbf{E}(x,t)=E_0\sin(kx-\omega t)\,\hat{z}$ with $E_0=cB_0\approx3\times10^8\times10^{-5}=3\times10^3\,$V/m, $\omega=2\pi\times10^{10}\,$s^{-1}, $k=\omega/c$.

Q.5If the relative permeability and relative permittivity of a medium are 1.0 and 2.25 respectively, find the speed of the electromagnetic wave in this medium.v
Solution

In a medium $v=1/\sqrt{\mu\varepsilon}=c/\sqrt{\mu_r\varepsilon_r}$. With $\mu_r=1.0$, $\varepsilon_r=2.25$, $v=c/\sqrt{2.25}=c/1.5=(3\times10^8)/1.5=2.0\times10^8\,$m/s.

Answer:

Speed $v=2.0\times10^{8}\,$m s^{-1}

Need a step-by-step explanation?

Use the AI Doubt Solver for formulas, derivations, and numericals you want explained in simpler English.

Ask AI Doubt
Back
Ch.4 ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT
Next
Ch.6 Ray Optics