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Electromagnetic Waves MCQs for NEET — Physics Questions with Answers

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A flood light is covered with a fitter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave

$E_{x} = 36 \sin (1.20 \times 10^{7} z - 3.6 \times 10^{15} t) V / m$

The average intensity of beam in W/ $m^{2}$ will be

6.88

3.44

1.72

0.86

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Explanation

$I n t e n s i t y = (\frac{1}{2} ε_{0} {E_{0}}^{2}) C = \frac{1}{2} \times 8.85 \times 10^{- 12} \times {(36)}^{2} \times 3 \times 10^{8} = 1.72 W / m^{2}$

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The average energy - density of electromagnetic wave given by E = (50 N/C) sin( $ω t - k x)$ will be nearly

$10^{- 8} J / m^{3}$

$10^{- 7} J / m^{3}$

$10^{- 6} J / m^{3}$

$10^{- 5} J / m^{3}$

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Explanation

$E = 50 \sin (ω t - k x) A v g . e n e r g y d e n s i t y = \frac{1}{2} ε_{0} {E^{2}}_{0} = \frac{1}{2} \times 8.85 \times 10^{- 12} \times {(50)}^{2} = 1.1 \times 10^{- 8} J / m^{3}$

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A larger parallel plate capacitor, whose plates have an area of 1 $m^{2}$ are separated from each other by 1 mm, is being charged at a rate of 25.8 V/s. If the plates has dielectric constant 10, then the displacement current at this instant is

25 $μ A$

11 $μ A$

2.2 $μ A$

1.1 $μ A$

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Explanation

During charging of capacitor,

Conduction current = Displacement current

$\frac{d Q}{d t} = i_{d} Q = C V i_{d} = C \frac{d V}{d t} = (\frac{K A ε_{0}}{d}) \times \frac{d V}{d t} = \frac{10 \times 1 \times 8.85 \times 10^{- 12}}{10^{- 3}} \times 25.8 = 2.28 μ A$

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A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. The displacement current through this area is

i/2

i/4

i/8

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Explanation

At any instant, let charge on capacitor be Q

Electric field between plates, $\vec{E} = \frac{Q}{A ε_{0}}$

Flux through area $\frac{A}{2}, ϕ_{E} = \frac{E A}{2}$

Displacement current ,

$i_{d} = ε_{0} \frac{d ϕ_{E}}{d t} = \frac{1}{2} \frac{d Q}{d t} = \frac{i}{2}$

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The sun delivers $10^{4} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions (10 $\times$ 10) $m^{2}$ will be

$10^{4}$ W

$10^{5} W$

$10^{6} W$

$10^{7} W$

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Explanation

$P = I n t e n s i t y \times A r e a = 10^{4} \times 10 \times 10 = 10^{6} W$

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The sun delivers $10^{3} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions 8 m $\times$ 20 m will be

$2.56 \times 10^{4} W$

$6.4 \times 10^{5} W$

$4.0 \times 10^{5} W$

1.6 $\times 10^{5} W$

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Explanation

$P = I n t e n s i t y \times A r e a = 10^{3} \times 8 \times 20 = 1.6 \times 10^{5} W$

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The wave of wavelength 5900 $\overset{0}{A}$ emitted by any atom or molecule must have some finite total length which is known as coherence length. For sodium light, this length is 2.4 cm. The number of oscillations in this length will be

$4.068 \times 10^{8}$

$4.068 \times 10^{4}$

$4.068 \times 10^{6}$

$4.068 \times 10^{5}$

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Explanation

No. of oscillation = Wavelengths in coherence length

$n = \frac{2.4 c m}{5900 \overset{0}{A}} = 4.068 \times 10^{4}$

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A radio wave has a maximum electric field intensity of $10^{- 4} V m^{- 1}$ on arrival at a receiving antenna. The maximum magnetic flux density of such a wave is

zero

$3 \times 10^{4} T$

$5.8 \times 10^{- 9} T$

$3.3 \times 10^{- 13} T$

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Explanation

Magnetic flux density, $B = \frac{ϕ_{B}}{A}$

$B_{0} = \frac{E_{0}}{C} = \frac{10^{- 4}}{3 \times 10^{8}} = 3.3 \times 10^{- 13} T$

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The transmitting antenna of a radio station is mounted vertically. At a point 10 km due north of the transmitter , the peak electric field is $10^{- 3}$ V/m. The amplitude of the radiated magnetic field is

$3.33 \times 10^{- 10} T$

$3.33 \times 10^{- 12} T$

$10^{- 3} T$

$3 \times 10^{5} T$

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Explanation

$E_{0} = 10^{- 3} V / m B_{0} = \frac{E_{0}}{C} = \frac{10^{- 3}}{3 \times 10^{8}} = 3.33 \times 10^{- 12} T$

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A plane EM-wave of wave intensity of 10 $W / m^{2}$ strikes a small mirror of area 20 $c m^{2}$ , held perpendicular to the approaching wave. The radiation force on the mirror will be

$6.6 \times 10^{- 11} N$

$1.33 \times 10^{- 11} N$

$1.33 \times 10^{- 10} N$

$6.6 \times 10^{- 10} N$

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Explanation

For perfectly reflecting surface,

$F = \frac{2 U_{a v}}{C} [U_{a v} = A v g . e n e r g y d e n s i t y] = \frac{2 [I A]}{C} = \frac{2 \times 10 \times 20 \times 10^{- 4}}{3 \times 10^{8}} = 1.33 \times 10^{- 10} N$

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