Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
An electromagnetic wave travelling in $x$-direction is described by field equation $E_y = 300 \sin \omega \left(t - \frac{x}{c}\right)$. If the electron is restricted to move in $y$-direction only with speed of $1.5 \times 10^6$ m/s then ratio of maximum electric and magnetic forces acting on the electron is _______.
(1) 200
(2)
(3)
(4)
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Answer: Option (1)
Key idea: Forces on charged particle in electromagnetic wave; Relation between E and B in EM waves
From the given equation, electric field amplitude is $E_0 = 300$ V/m
For an EM wave, magnetic field amplitude is $B_0 = E_0/c = 300/(3 \times 10^8) = 10^{-6}$ T
Maximum electric force on electron: $F_{E,max} = eE_0 = 300e$ N
Since electron moves in y-direction with speed $v = 1.5 \times 10^6$ m/s and $\vec{B}$ is perpendicular, maximum magnetic force: $F_{B,max} = evB_0 = e \times 1.5 \times 10^6 \times 10^{-6} = 1.5e$ N
An electromagnetic wave travels in free space along the $x$-direction. At a particular point in space and time, $\vec{B} = 2 \times 10^{-7}$ $\hat{j}$ T is associated with this wave. The value of corresponding electric field $\vec{E}$ at this point is ________ V/m.
(1) $60\hat{k}$
(2) $-60\hat{k}$
(3) $30\hat{k}$
(4) $-600\hat{k}$
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Answer: Option (2)
Key idea: Electromagnetic wave propagation: relationship between E, B, and direction of propagation
For an EM wave in free space, the magnitude relation is $E = cB$ where $c = 3 imes 10^8$ m/s
A magnetic field vector in an electromagnetic wave is represented by $\vec{B} = B_0 \sin\left(2\pi vt - \frac{2\pi x}{\lambda}\right) \hat{j}$. Its associated electric field vector is ________.
Key idea: Electromagnetic wave properties: E and B field relationship, direction of propagation, and magnitude relation E₀ = cB₀
Identify that the wave propagates in $+\hat{i}$ direction from the phase term $(2\pi vt - \frac{2\pi x}{\lambda})$
Note that $\vec{B} = B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{j}$ is in the $\hat{j}$ direction
Use $\vec{E} \times \vec{B}$ must point in direction of propagation ($\hat{i}$). Since $\vec{B}$ is along $\hat{j}$, $\vec{E}$ must be along $-\hat{k}$ to satisfy $(-\hat{k}) \times \hat{j} = +\hat{i}$
Apply the magnitude relationship $E_0 = cB_0$ where $c = v\lambda$ (speed of light = frequency × wavelength)
Substitute to get $E_0 = v\lambda B_0$
Write the complete electric field: $\vec{E} = -v\lambda B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{k}$
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The electromagnetic wave exerts pressure on the surface on which they are allowed to fall.
Reason (R) : There is no mass associated with the electromagnetic waves.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both (A) and (R) are true and (R) is the correct explanation of (A)
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) (A) is true but (R) is false
(4) (A) is false but (R) is true
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Answer: Option (2)
Key idea: Radiation pressure and momentum of electromagnetic waves
**Assertion (A)**: EM waves exert pressure on surfaces - This is **TRUE**. Radiation pressure $P = I/c$ (absorption) or $P = 2I/c$ (reflection)
**Reason (R)**: EM waves have no mass - This is **TRUE**. Photons have zero rest mass
**Key Question**: Does (R) explain (A)?
EM waves exert pressure because they carry **momentum** $p = E/c = h/\lambda$, not because they lack mass
The momentum transfer (despite zero mass) causes the pressure, so (R) does NOT correctly explain (A)
**Conclusion**: Both statements are true, but (R) is not the correct explanation of (A)
A point light source emits E.M. waves in free space. A detector, placed at a distance of $L$ m, measures the intensity as $I_o$. The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as $45°$. The measured intensity at new location will be _______.
(1) $\frac{I_o}{4}$
(2) $I_o$
(3) $\frac{I_o}{\sqrt{2}}$
(4) $\frac{I_o}{2}$
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Answer: Option (2)
Key idea: Intensity of electromagnetic waves from a point source depends only on distance, not direction. For isotropic point sources, intensity is uniform on spherical surfaces.
For a point source, intensity is given by $I = \frac{P}{4\pi r^2}$ where $P$ is power and $r$ is distance
At original location: distance = $L$, so $I_1 = \frac{P}{4\pi L^2} = I_o$
Detector moves to new location on the SAME spherical surface, so distance remains $L$
At new location: distance = $L$, so $I_2 = \frac{P}{4\pi L^2} = I_o$
The $45°$ angle is irrelevant - intensity depends only on distance for isotropic point sources
Therefore, measured intensity at new location = $I_o$
For an electromagnetic wave propagating through vacuum, $\vec{k}$, $\vec{E}$ and $\omega$ represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by :
(1) $\frac{\vec{E} \times \vec{k}}{\omega}$
(2)
(3)
(4)
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Answer: Option (1)
Key idea: Electromagnetic wave relations: magnetic field in terms of electric field and propagation vector
For an EM wave in vacuum, $\vec{E}$, $\vec{B}$, and $\vec{k}$ are mutually perpendicular and form a right-handed system
The magnitudes are related by $B = \frac{E}{c}$ where $c$ is the speed of light
For a wave, $k = \frac{\omega}{c}$, so $c = \frac{\omega}{k}$
The magnetic field must be perpendicular to both $\vec{E}$ and $\vec{k}$, suggesting a cross product
Using the relation $\vec{B} = \frac{1}{\omega}(\vec{k} \times \vec{E})$ or equivalently $\vec{B} = \frac{1}{\omega}(\vec{E} \times \vec{k})$ with appropriate sign
The answer is $\vec{B} = \frac{\vec{E} \times \vec{k}}{\omega}$
Consider the equation $H = \frac{x^p e^q E^r}{t^s}$
Where $H =$ magnetic field; $E =$ electric field, $\epsilon =$ permittivity, $x =$ distance, $t =$ time
The values of $p, q, r$ and $s$ respectively are:
(1) $1, 1, 1, 1$
(2) $-1, 1, 2, 1$
(3) $1, -1, -2, 1$
(4) $-1, -2, -2, 1$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Dimensional Analysis of Electromagnetic Quantities
A monochromatic source of light operating at 15 kW emits $2.5 \times 10^{22}$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take $h = 6.6 \times 10^{-34}$ J.s and $c = 3 \times 10^8$ m/s).
(1) Microwave
(2) Infrared
(3) Visible
(4) Ultraviolet
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Answer: Option (4)
Key idea: Photon energy and wavelength calculation to identify electromagnetic spectrum region
