Question 4

   Question: The property which is not of an electromagnetic wave travelling in free space is that:

   Options: 

       A. They are transverse in nature

       B. The energy density in electric field is equal to energy density in magnetic field

       C. They travel with a speed equal to $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$

       D. They originate from charges moving with uniform speed

   Correct Answer: D

   Year: [NEET 2024]

   Solution: The EM waves originate from an accelerating charge. The charge moving with uniform velocity produces steady state magnetic field.

   Step Solution:

    1.  Recall that EM waves are transverse, meaning $\vec{E}$ and $\vec{B}$ are perpendicular to the direction of propagation.

    2.  Recall the speed of light in a vacuum is defined by the constants $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

    3.  Consider energy distribution: The energy density is shared equally between the electric and magnetic fields.

    4.  Identify the source: Static charges create electric fields; uniform motion creates steady magnetic fields.

    5.  Conclusion: Only an accelerated or oscillating charge creates time-varying fields that propagate as a wave.

   Difficulty Level: Easy

   Concept Name: Production and Properties of EM Waves

   Short cut solution: Remember that "Acceleration = Radiation." Uniform speed cannot produce a wave.

Question 20

   Question: Out of the following options which one can be used to produce a propagating electromagnetic wave?

   Options: 

       A. A chargeless particle

       B. An accelerating charge

       C. A charge moving at constant velocity

       D. A stationary charge

   Correct Answer: B

   Year: 2016 NEET Phase-1

   Solution: An accelerating charge is used to produce oscillating electric and magnetic fields, hence the electromagnetic wave.

   Step Solution:

    1.  Evaluate a stationary charge: It produces only a static electric field.

    2.  Evaluate a chargeless particle: It cannot interact with or produce electric/magnetic fields.

    3.  Evaluate a charge at constant velocity: It produces a steady magnetic field, but the fields do not oscillate or propagate.

    4.  Evaluate an accelerating charge: It creates a time-varying electric field, which induces a time-varying magnetic field.

    5.  Apply Maxwell's Theory: These mutually regenerating, time-varying fields propagate through space as an EM wave.

   Difficulty Level: Easy

   Concept Name: Source of Electromagnetic Waves

   Short cut solution: Only an accelerating charge creates the required oscillating fields for wave propagation.

Question 31

   Question: Which of the following statement is false for the properties of electromagnetic waves?

   Options: 

       A. Both electric and magnetic field vectors attain the maxima and minima at the same place and same time.

       B. The energy in electromagnetic wave is divided equally between electric and magnetic vectors.

       C. Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave.

       D. These waves do not require any material medium for propagation.

   Correct Answer: C

   Year: Not specified in the source (appears between 2011 and 2010 Mains).

   Solution: In an electromagnetic wave both electric and magnetic vectors are perpendicular to each other as well as perpendicular to the direction of propagation of wave.

   Step Solution:

    1.  Examine Phase (Option A): $\vec{E}$ and $\vec{B}$ are in phase, meaning they reach maxima/minima simultaneously. (True)

    2.  Examine Energy (Option B): Energy density $u_E = \frac{1}{2}\epsilon_0 E^2$ and $u_B = \frac{B^2}{2\mu_0}$ are equal. (True)

    3.  Examine Medium (Option D): EM waves are non-mechanical and travel through vacuum. (True)

    4.  Examine Vector Orientation (Option C): $\vec{E}$ and $\vec{B}$ must be mutually perpendicular ($\vec{E} \perp \vec{B}$), not parallel.

    5.  Conclusion: Statement C is false because it claims the fields are parallel.

   Difficulty Level: Easy

   Concept Name: Characteristics of Electromagnetic Waves

   Short cut solution: In an EM wave, everything is perpendicular: $\vec{E} \perp \vec{B} \perp \text{Direction of Propagation}$.

Question 34

   Question: The velocity of electromagnetic radiation in a medium of permittivity $\epsilon_0$ and permeability $\mu_0$ is given by

   Options: 

       A. $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$

       B. $\sqrt{\mu_0 \epsilon}$

       C. $\sqrt{\epsilon_0 \mu}$

       D. $\sqrt{\mu_0 \epsilon_0}$

   Correct Answer: A

   Year: 2008

   Solution: The velocity of electromagnetic radiation in vacuum is $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$; where $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of vacuum.

   Step Solution:

    1.  Identify the medium as free space/vacuum based on the constants $\epsilon_0$ and $\mu_0$.

    2.  Recall Maxwell’s prediction that the speed of electromagnetic waves is determined by the electromagnetic properties of the medium.

    3.  Use the relation derived from the wave equation: $v^2 = \frac{1}{\mu \epsilon}$.

    4.  Substitute the vacuum constants $\mu_0$ and $\epsilon_0$ into the relation.

    5.  Calculate the final velocity formula: $v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

   Difficulty Level: Easy

   Concept Name: Speed of Light in Vacuum / Maxwell’s Equations

   Short cut solution: This is a fundamental constant formula; remember $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

Question 35

   Question: The electric and magnetic field of an electromagnetic wave are

   Options: 

       A. in opposite phase and perpendicular to each other

       B. in opposite phase and parallel to each other

       C. in phase and perpendicular to each other

       D. in phase and parallel to each other.

   Correct Answer: C

   Year: 2007, 1994

   Solution: In electromagnetic wave, electric and magnetic field are in phase and perpendicular to each other and also perpendicular to the direction of the propagation of the wave.

   Step Solution:

    1.  Analyze the wave structure: EM waves are transverse, meaning fields oscillate perpendicular to the direction of travel.

    2.  Examine the spatial relationship: By Maxwell's equations, the oscillating electric field $\vec{E}$ and magnetic field $\vec{B}$ are mutually perpendicular ($\vec{E} \perp \vec{B}$).

    3.  Examine the temporal relationship: The fields are generated by the same source and propagate together.

    4.  Determine phase: Because they reach their maximum and minimum values at the same time and place, they are said to be in phase.

    5.  Conclusion: The fields are both in phase and perpendicular.

   Difficulty Level: Easy

   Concept Name: Transverse Nature and Phase of EM Waves

   Short cut solution: Remember "PIP": Perpendicular to each other, In-phase, and Perpendicular to propagation.

Question 38

   Question: The electric field associated with an em wave in vacuum is given by $\vec{E} = \hat{i} 40 \cos(kz - 6 \times 10^8 t)$, where $E, z$ and $t$ are in volt/m, meter and seconds respectively. The value of wave vector $k$ is.

   Options: 

       A. $2 \text{ m}^{-1}$

       B. $0.5 \text{ m}^{-1}$

       C. $6 \text{ m}^{-1}$

       D. $3 \text{ m}^{-1}$

   Correct Answer: A

   Year: 2012

   Solution: Compare the given equation with $E = E_0 \cos(kz - \omega t)$, we get, $\omega = 6 \times 10^8 \text{ s}^{-1}$. Wave vector, $k = \frac{\omega}{c} = \frac{6 \times 10^8 \text{ s}^{-1}}{3 \times 10^8 \text{ ms}^{-1}} = 2 \text{ m}^{-1}$.

   Step Solution:

    1.  Identify Equation: Write down the standard plane wave equation: $E = E_0 \cos(kz - \omega t)$.

    2.  Extract Frequency: Compare the given equation $\vec{E} = \hat{i} 40 \cos(kz - 6 \times 10^8 t)$ to find angular frequency $\omega = 6 \times 10^8 \text{ rad/s}$.

    3.  State Relationship: Use the fundamental wave parameter relationship: $k = \frac{\omega}{c}$, where $c$ is the speed of light in vacuum ($3 \times 10^8 \text{ m/s}$).

    4.  Substitution: Substitute the values into the formula: $k = \frac{6 \times 10^8}{3 \times 10^8}$.

    5.  Calculate: Divide to find the wave vector: $k = 2 \text{ m}^{-1}$.

   Difficulty Level: Easy

   Concept Name: Wave Vector and Angular Frequency Relationship

   Short cut solution: The wave vector $k$ is simply the coefficient of $t$ divided by $c$ ($k = \omega/c$). $6/3 = 2$.

Question 39

   Question: The velocity of electromagnetic wave is parallel to.

   Options: 

       A. $\vec{B} \times \vec{E}$

       B. $\vec{E} \times \vec{B}$

       C. $\vec{E}$

       D. $\vec{B}$

   Correct Answer: B

   Year: 2002

   Solution: According to Maxwell, the electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vectors at right angles to each other as well as at right angles to the direction of wave propagation. Therefore, the velocity of electromagnetic wave is parallel to $\vec{E} \times \vec{B}$.

   Step Solution:

    1.  Recall Wave Nature: Identify that EM waves consist of oscillating electric ($\vec{E}$) and magnetic ($\vec{B}$) fields.

    2.  Define Transversality: In an EM wave, $\vec{E}$ and $\vec{B}$ are always perpendicular to each other and to the direction of propagation.

    3.  Vector Calculus Application: The direction of energy flow (Poynting vector) and wave propagation is determined by the cross product of the fields.

    4.  Directional Rule: Using the right-hand rule, if $\vec{E}$ is on the $y$-axis and $\vec{B}$ is on the $z$-axis, propagation is along the $x$-axis ($\hat{j} \times \hat{k} = \hat{i}$).

    5.  Conclusion: The velocity vector $\vec{v}$ must therefore be parallel to the resultant vector of $\vec{E} \times \vec{B}$.

   Difficulty Level: Easy

   Concept Name: Direction of Propagation of EM Waves

   Short cut solution: Alphabetical order: E comes before B, so the cross product is $\vec{E} \times \vec{B}$.