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The magnetic permeability of free space ...

The magnetic permeability of free space is …………………… .

Text Solution

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`mu_0 = 4 pi xx 10^(-7) T mA^(-1)`
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If epsilon_0 and mu_o are, respectively, the electric permittivity and magnetic permeability of free space, epsilon and mu the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is….

If epsilon_(0) and mu_(0) are, respectively, the electric permittivity and magnetic permeability of free space, epsilon and mu the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is "________________" .

Knowledge Check

  • If epsi_(0) and mu_(0) are respectively, the electric permittivity and the magnetic permeability of free space, epsi and mu the corresponding quantities in a medium, the refractive index of the medium is

    A
    `sqrt((mu epsi)/(mu_(0) epsi_(0)))`
    B
    `(mu epsi)/(mu_(0) epsi_(0))`
    C
    `sqrt((mu_(0) epsi_(0))/(mu epsi))`
    D
    `sqrt((mu mu_(0))/(epsi epsi_(0)))`
  • If epsilon_0 and mu_0 are respectively the electric permittivity and the magnetic permeability of free space and epsilon and mu the corresponding quantities in a medium, the refractive index of the medium is

    A
    (a) `sqrt(((muepsilon)/(mu_0epsilon_0)))`
    B
    (b) `(muepsilon)/(mu_0epsilon_0)`
    C
    (c) `sqrt(((mu_0epsilon_0)/(muepsilon)))`
    D
    (d) `sqrt(((mumu_0)/(epsilonepsilon_0)))`
  • If mu_o represents the magnetic permeability constant in free space and Sigma_o is the permitivity in vacuum, and C the speed of light in vacuum , then

    A
    `Sigma_o=sqrt(mu_oC)`
    B
    `Sigma_o^(-2)=mu_oC^(-1)`
    C
    `Sigma_o^(-1)=mu_o^(-1)C^(-2)`
    D
    `Sigma_o=mu_o^(-1) C^(-2)`
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    Magnetic permeability of ferromagnetic substance is