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The speed of electromagnetic waves in va...

The speed of electromagnetic waves in vacuum is -

(A) `sqrt(in_0mu_0)`
(B) 1/`sqrt(mu_0in_0)`
(C) `in_0mu_0`
(D) 1/`(in_0mu_0)`

A

`sqrt(in_0mu_0)`

B

1/`sqrt(mu_0in_0)`

C

`in_0mu_0`

D

1/`(in_0mu_0)`

Text Solution

Verified by Experts

The correct Answer is:
B
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Electromagnetic waves propagate through free space or a medium as transverse waves. The electric and magnetic fields are perpendicular to each other as well as perpendicular to the direction of propagation of waves at each point. In the direction of wave propagation, electric field vecE and magnetic field vecB form a right-handed cartesian coordinate system. During the propagation of electromagnetic wave, total energy of electromagnetic wave is distributed equally between electric and magnetic fields. Since in_0 and mu_0 are permittivity and permeability of free space, the velocity of electromagnetic wave, c=(in_0 mu_0)^(-1//2) . Energy density i.e., energy in unit volume due to electric field at any point, u_E=1/2in_0E^2 Similarly, energy density due to magnetic field , u_M=1/(2mu_0)B^2 . If the electromagnetic wave propagates along x-direction, then the equations of electric and magnetic field are respectively. E=E_0sin(omegat-kx) and B=B_0sin(omegat-kx) Here, the frequency and the wavelength of oscillating electric and magnetic fields are f=omega/(2pi) and lambda=(2pi)/k respectively. Thus E_"rms"=E_0/sqrt2 and B_"rms"=B_0/sqrt2 , where E_0/B_0=c . Therefore, average energy density baru_E=1/2in_0E_"rms"^2 and baru_M=1/(2mu_0)B_"rms"^2 . The intensity of the electromagnetic wave at a point, I=cbaru=c(baru_E+baru_B) . To answer the following questions , we assume that in case of propagation of electromagnetic wave through free space, c=3xx10^8 m.s^(-1) and mu_0=4pixx10^(-7) H.m^(-1) If the peak value of electric field at a point in electromagnetic wave is 15 V . m^(-1) , then average electrical energy density (in j . m^(-3) )

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