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Electromagnetic Waves MCQs for NEET — Physics Questions with Answers

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In a region of free space, the electric field at some instant of time is $\vec{E} = (80 \hat{i} + 32 \hat{j} - 64 \hat{k})$ V/m and the magnetic field is $\vec{B} = (0.2 \vec{i} + 0.08 \hat{j} + 0.29 \hat{k}) μ T$ . The Poynting vector for these fields is

$- 11.52 \hat{i} + 28.8 \hat{j}$

$- 28.8 \hat{i} + 11.52 \hat{i}$

$28.8 \hat{i} - 11.52 \hat{j}$

$11.52 \hat{i} - 28.8 \hat{j}$

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Explanation

$Pointing vector = (\vec{E} \times \vec{B}) \frac{1}{μ_{0}} = \frac{10^{- 6}}{μ_{0}} |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 80 & 32 & - 64 \\ 0.2 & 0.08 & 0.29 \end{matrix}| = 11.52 \hat{i} - 28.8 \hat{j}$

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A plane electromagnetic wave propagating in the x-direction has wavelength of 60 mm. The electric field is in the y-direction and its maximum magnitude is 33 V/ $m^{- 1}$ . The equation for the electric field as function of x and t is

11 sin $π$ (t - x/c)

33 sin $π \times 10^{10}$ (t - x/c)

33 sin $π$ (t - x/c)

11 sin $π \times 10^{11}$ (t - x/c)

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Explanation

$λ = 60 m m ω = 2 π f = 2 π (\frac{c}{λ}) = 2 π \times \frac{3 \times 10^{8}}{60 \times 10^{- 3}} = π \times 10^{10} rad / s E = E_{0} \sin [ω (t - x / c)] = 33 \sin [π \times 10^{10} (t - x / c)]$

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In an electromagnetic wave, the electric field oscillated sinusoidally with amplitude 48 V $m^{- 1}$ , the RMS value of the oscillating magnetic field will be :

$1.6 \times 10^{- 8} T$

$1.6 \times 10^{- 9} T$

$144 \times 10^{- 8} T$

$11.3 \times 10^{- 8} T$

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Explanation

$B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T B_{rms} = \frac{B_{0}}{\sqrt{2}} = \frac{1.6 \times 10^{- 7}}{\sqrt{2}} = 11.3 \times 10^{- 8} T$

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The magnetic field in the plane electromagnetic wave is given by

$B_{z} = 2 \times 10^{- 7} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) t e s l a$

The expression for the electric field will be

$E_{z} = 30 \sqrt{2} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) V / m$

$E_{z} = 60 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) V / m$

$E_{y} = 30 \sqrt{2} \sin (0.5 \times 10^{11} x + 1.5 \times 10^{3} t) V / m$

$E_{y} = 60 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) V / m$

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Explanation

$\vec{B}$ along +z direction

EM wave travels along -x direction

Direction of $\vec{E} \times \vec{B}$ gives direction of wave

$∴ \vec{E}$ along y direction

Amplitude of electric field, $E_{0}$

$E_{0} = B_{0} c = 2 \times 10^{- 7} \times 3 \times 10^{8} = 60 V / m ∴ Electric field . E_{y} = 60 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t)$

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For a transparent medium relative permeability and permittivity, $μ_{r} a n d ε_{r}$ are 1.0 and 1.44 respectively. The velocity of light in this medium would be:

2.5 $\times 10^{8}$ m/s

3 $\times 10^{8}$ m/s

2.08 $\times 10^{8}$ m/s

4.32 $\times 10^{8}$ m/s

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Explanation

$\begin{array}{l} μ = μ_{r} μ_{0} \\ ε = ε_{r} ε_{0} \\ v = \frac{1}{\sqrt{με}} \\ c = \frac{1}{\sqrt{μ_{0} ε_{0}}} \\ ∴ v = \frac{c}{\sqrt{μ_{r} ε_{r}}} = 2.5 \times 10^{8} m / s \end{array}$

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A star emitting light of wavelength 5896 Å is moving away from the earth with a speed of 3600 km/sec. The wavelength of light observed on earth will

Decrease by 5825.25 Å

Increase by 5966.75 Å

Decrease by 70.75 Å

Increase by 70.75 Å

(c = 3 × 10⁸ m/sec is the speed of light)

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Explanation

$ν' = ν \sqrt{(\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}})} λ = λ \sqrt{(\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}})} = λ \sqrt{(\frac{1 + \frac{3600}{c}}{1 - \frac{3600}{c}})} = λ \sqrt{\frac{1.012}{0.988}} = 1.012 λ = 70.75 \overset{o}{A}$

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A heavenly body is receding away from the earth such that the fractional change in λ is 1, then its velocity is :

$\frac{3 C}{5}$

$\frac{C}{5}$

$\frac{2 C}{5}$

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Explanation

$W h e n a p l a n e t r e c e d e s a w a y f r o m t h e e a r t h a n d f i s t h e f r e q u e n c y o f v i b r a t i o n : f = \frac{c}{λ} I f v = v e l o c i t y o f m o v i n g a w a y f r o m t h e e a r t h a n d λ' = a p p a r e n t w a v e l e n g t h t o a n o b e s e r v e r o n t h e e a r t h; λ' = \frac{c + v}{f} \Rightarrow λ' = (\frac{c + v}{c}) λ F r a c t i o n a l c h a n g e i n w a v e l e n t h = \frac{λ' - λ}{λ} = \frac{(\frac{c + v}{c}) λ - λ}{λ} = \frac{v}{c} = 1 \Rightarrow v = c$

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In the propagation of electromagnetic waves, the angle between the direction of propagation and plane of polarisation is

0^o

45^o

90^o

180^o

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Explanation

Plane of polarization is a confinement of the electric/magnetic field vector to a given plane along the direction of propagation. Therefore the angle between them is $0^{0}$

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A light has amplitude A and the angle between analyzer and polariser is 60°. Light reflected by analyzer has amplitude

$A \sqrt{2}$

$A / \sqrt{2}$

$\sqrt{3} A / 2$

A/2

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Explanation

The amplitude will be $A \cos 60^{o} = A / 2$

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Which of the following conditions is necessary for the production of electromagnetic waves?

A. Stationary charges B. Charges in uniform motion (steady currents) C. Accelerated charges D. Charges at rest in a uniform electric field

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Explanation

According to Maxwell's theory, accelerated charges radiate electromagnetic waves. Neither stationary charges nor charges in uniform motion can be sources of electromagnetic waves. (Context: 'It is an important result of Maxwellâ€™s theory that accelerated charges radiate electromagnetic waves.')

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