Modern Physics 7 Question 71

1. A plane electromagnetic wave having a frequency v=23.9 GHz propagates along the positive z-direction in free space. The peak value of the electric field is 60V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave?

(a) B=2×107sin(0.5×103z+1.5×1011t)i^

(Main 2019, 12 April II)

(b) B=2×107sin(0.5×103z1.5×1011t)i^

(c) B=60sin(0.5×103x+1.5×1011t)k^

(d) B=2×107sin(1.5×102x+0.5×1011t)j^

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Solution:

  1. In an electromagnetic wave, magnetic field and electric field are perpendicular to each other and both are also perpendicular to the direction of propagation of wave.

Now, given direction of propagation is along z-direction. So, magnetic field is in either x or y direction. Also, angular wave number for wave is

k=2πλ=2πνc=2π×23.9×1093×1080.5×103m1

and angular frequency ω for wave is

ω=2πν=2π×23.9×109Hz=1.5×1011Hz

Magnitude of magnetic field is

B0=E0c=603×108=2×107T

As the general equation of magnetic field of an electromagnetic wave propagating in +z-direction is given as,

B=B0sin(kzωt)i^ or j^

Thus, substituting the values of B0,k and ω, we get B=2×107sin(0.5×103z1.5×1011t)i^ or j^



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