Home
Class 12
PHYSICS
If dielectric constant and relative magn...

If dielectric constant and relative magnetic permeability of a medium are 4 and 2.25 receptively, then the velocity of electromagnetic wave in that medium is `nxx10^8 m.s^(-1)` Find the value of n.

Text Solution

Verified by Experts

The correct Answer is:
1
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROMAGNETIC WAVES

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE|15 Videos

  • ELECTROMAGNETIC WAVES

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|17 Videos

  • ELECTROMAGNETIC WAVES

    CHHAYA PUBLICATION|Exercise ENTRANCE CORNER (COMPREHENSION )|7 Videos

  • ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|28 Videos

  • ELECTROMAGNETISM

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|26 Videos

Similar Questions

Explore conceptually related problems

When the frequency of an electromagnetic wave and the amplitude of its associated magnetic field at a point are 10^8 Hz and 10^(-10) T respectively, then the amplitude of electric field be nxx10^(-2) V.m^(-1) . Find the value of n.

The relative magnetic permeability of two media are 1.0004 and 1.00005 . Ration of their magnetic susceptibility is n :1 . Find n .

Knowledge Check

  • Statement I: The ratio of the amplitudes of electric and magnetic fields at a point in the electromagnetic wave is same as the velocity of wave. Statement II: If in a medium, mu and in are the magnetic permeability and the electric permittivity respectively, then 1//sqrt(muin) is the velocity of electromagnetic wave in that medium.

    A
    Statement I is true, statement II is true, statement II is a correct explanation for statement I.
    B
    Statement I is true, statement II is true, statement II is not a correct explanation for statement I.
    C
    Statement I is true, statement II is false.
    D
    Statement I is false, statement II is true.

  • 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) )

    A
    `4.48xx10^(-9)`
    B
    `9.95xx10^(-9)`
    C
    `4.98xx10^(-10)`
    D
    `9.95xx10^(-10)`

  • 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) The peak value of magnetic field (in Wb . m^(-2) ) at that point

    A
    `5xx10^(-8)`
    B
    `45xx10^(-8)`
    C
    `5xx10^8`
    D
    `45xx10^8`

    • Similar Questions

    Explore conceptually related problems

    If the velocity of light in a medium is 2xx10^9 m//s What will be the refractive index of that medium?

    The velocity of light in a medium is 2 xx 10^8m//s . What will be the refractive index of the medium?

    How is the velocity of an electromagnetic wave in a medium related with the electric and the magnetic properties of that medium?

    Due to the sunlight, if the amplitude of the electric field at the earth surface is 900 V . m^(-1) , then the amplitude of the magnetic field is nxx10^(-6) T. Find the value of n.

    A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9xx10^(-7)m^(2) . If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140rad/s. If the Young's modulus of the material of the wire is nxx10^(9)N*m^(-2) , then find out the value of n.

    Home
    Profile