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Electromagnetic Waves question

2018 · Shift 1 · Q50
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Electromagnetic Waves question

2018 · Shift 1 · Q50

JEE AdvancedPhysicsElectromagnetic WavesMCQ+3 / −1
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E][E][E] and [B][B][B] stand for dimensions of electric and magnetic fields respectively, while [ε0]\left[ {{\varepsilon _0}} \right][ε0​] and [μ0]\left[ {{\mu _0}} \right][μ0​] stand for dimensions of the permittivity and permeability of free space respectively. [L]\left[ L \right][L] and [T]\left[ T \right][T] are dimensions of length and time respectively. All the quantities are given in SISISI units. The relation between [ε0]\left[ {{\varepsilon _0}} \right][ε0​] and [μ0]\left[ {{\mu _0}} \right][μ0​] is
  1. A
    [μ0]=[ε0][L]2[T]−2\left[ {{\mu _0}} \right] = \left[ {{\varepsilon _0}} \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}}[μ0​]=[ε0​][L]2[T]−2
  2. B
    [μ0]=[ε0][L]−2[T]2\left[ {{\mu _0}} \right] = \left[ {{\varepsilon _0}} \right]{\left[ L \right]^{ - 2}}{\left[ T \right]^2}[μ0​]=[ε0​][L]−2[T]2
  3. C
    [μ0]=[ε0]−1[L]2[T]−2\left[ {{\mu _0}} \right] = {\left[ {{\varepsilon _0}} \right]^{ - 1}}{\left[ L \right]^2}{\left[ T \right]^{ - 2}}[μ0​]=[ε0​]−1[L]2[T]−2
  4. D
    [μ0]=[ε0]−1[L]−2[T]2\left[ {{\mu _0}} \right] = {\left[ {{\varepsilon _0}} \right]^{ - 1}}{\left[ L \right]^{ - 2}}{\left[ T \right]^2}[μ0​]=[ε0​]−1[L]−2[T]2
View written solutionFree

Correct answer: D

Step-by-step Derivation

  1. Identify the fundamental relationship: The speed of light in a vacuum, denoted by c, is related to the permittivity of free space, ε₀, and the permeability of free space, μ₀, by the following equation from electromagnetic theory: c=1ε0μ0c = \frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}c=ε0​μ0​​1​

  2. Rearrange the equation: To find a direct relationship between ε₀ and μ₀, we can manipulate this equation. First, square both sides: c2=1ε0μ0{c^2} = \frac{1}{{{\varepsilon _0}{\mu _0}}}c2=ε0​μ0​1​

  3. Isolate μ₀: Now, rearrange the equation to express μ₀ in terms of ε₀ and c: μ0=1ε0c2{\mu _0} = \frac{1}{{{\varepsilon _0}{c^2}}}μ0​=ε0​c21​

  4. Perform dimensional analysis: We need to find the relationship between the dimensions of these quantities. Let [X] denote the dimension of a quantity X.

    • The dimension of permeability is [μ₀].
    • The dimension of permittivity is [ε₀].
    • The speed c has dimensions of length per time. So, [c]=[L][T]=[L][T]−1[c] = \frac{[L]}{[T]} = [L][T]^{-1}[c]=[T][L]​=[L][T]−1.
    • Therefore, the dimension of c² is [c2]=([L][T]−1)2=[L]2[T]−2[c^2] = ([L][T]^{-1})^2 = [L]^2[T]^{-2}[c2]=([L][T]−1)2=[L]2[T]−2.

  5. Substitute the dimensions into the equation: Now, substitute the dimensional forms back into the rearranged equation from Step 3: [μ0]=1[ε0][c2]\left[ {{\mu _0}} \right] = \frac{1}{{\left[ {{\varepsilon _0}} \right]\left[ {{c^2}} \right]}}[μ0​]=[ε0​][c2]1​ [μ0]=1[ε0][L]2[T]−2\left[ {{\mu _0}} \right] = \frac{1}{{\left[ {{\varepsilon _0}} \right]{{\left[ L \right]}^2}{{\left[ T \right]}^{ - 2}}}}[μ0​]=[ε0​][L]2[T]−21​

  6. Express the final relationship: To match the format of the options, we can bring the terms from the denominator to the numerator by inverting their exponents: [μ0]=[ε0]−1[L]−2[T]2\left[ {{\mu _0}} \right] = {\left[ {{\varepsilon _0}} \right]^{ - 1}}{\left[ L \right]^{ - 2}}{\left[ T \right]^2}[μ0​]=[ε0​]−1[L]−2[T]2

  7. Compare with the options:

    • A: [μ₀] = [ε₀][L]²[T]⁻² - Incorrect.
    • B: [μ₀] = [ε₀][L]⁻²[T]² - Incorrect.
    • C: [μ₀] = [ε₀]⁻¹[L]²[T]⁻² - Incorrect.
    • D: [μ₀] = [ε₀]⁻¹[L]⁻²[T]² - Correct.

Thus, the correct relation between the dimensions is given in option D.

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