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Electromagnetic Waves question

2015 · Shift 2 · Q45
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Electromagnetic Waves question

2015 · Shift 2 · Q45

JEE AdvancedPhysicsElectromagnetic WavesMultiple correct+4 / −2
In terms of potential difference V, electric current I, permittivity ε0{\varepsilon _0}ε0​, permeability μ0{\mu _0}μ0​ and speed of light c, the dimensionally correct equation(s) is(are) :
  1. A
    μ0I2=ε0V2{\mu _0}{I^2} = {\varepsilon _0}{V^2}μ0​I2=ε0​V2
  2. B
    ε0I=μ0V{\varepsilon _0}I = {\mu _0}Vε0​I=μ0​V
  3. C
    I=ε0cVI = {\varepsilon _0}cVI=ε0​cV
  4. D
    μ0cI=ε0V{\mu _0}cI = {\varepsilon _0}Vμ0​cI=ε0​V
View written solutionFree

Correct answer: A, C

  1. Write dimensions of the given quantities

We use base dimensions [M,L,T,I][M, L, T, I][M,L,T,I].

  • Electric current: [I]=I[I] = I[I]=I

  • Potential difference: [V]=workcharge=ML2T−2IT=ML2T−3I−1[V] = \frac{\text{work}}{\text{charge}} = \frac{ML^2T^{-2}}{IT} = ML^2T^{-3}I^{-1}[V]=chargework​=ITML2T−2​=ML2T−3I−1

  • Permittivity of free space ε0\varepsilon_0ε0​: From Coulomb's law, F=14πε0q2r2F = \frac{1}{4\pi \varepsilon_0}\frac{q^2}{r^2}F=4πε0​1​r2q2​ So, [ε0]=q2Fr2=(IT)2(MLT−2)L2=M−1L−3T4I2[\varepsilon_0] = \frac{q^2}{Fr^2} = \frac{(IT)^2}{(MLT^{-2})L^2} = M^{-1}L^{-3}T^4I^2[ε0​]=Fr2q2​=(MLT−2)L2(IT)2​=M−1L−3T4I2

  • Permeability of free space μ0\mu_0μ0​: Using c=1μ0ε0c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}c=μ0​ε0​​1​ we get [μ0]=[ε0]−1[c]−2[\mu_0] = [\varepsilon_0]^{-1}[c]^{-2}[μ0​]=[ε0​]−1[c]−2 Since [c]=LT−1,[c] = LT^{-1},[c]=LT−1, therefore [μ0]=(ML3T−4I−2)(L−2T2)=MLT−2I−2[\mu_0] = (ML^3T^{-4}I^{-2})(L^{-2}T^2) = MLT^{-2}I^{-2}[μ0​]=(ML3T−4I−2)(L−2T2)=MLT−2I−2

Also, [c]=LT−1[c] = LT^{-1}[c]=LT−1

  1. Check option A

μ0I2\mu_0 I^2μ0​I2 Dimension: [μ0I2]=(MLT−2I−2)(I2)=MLT−2[\mu_0 I^2] = (MLT^{-2}I^{-2})(I^2) = MLT^{-2}[μ0​I2]=(MLT−2I−2)(I2)=MLT−2

ε0V2\varepsilon_0 V^2ε0​V2 Dimension: [ε0V2]=(M−1L−3T4I2)(ML2T−3I−1)2[\varepsilon_0 V^2] = (M^{-1}L^{-3}T^4I^2)(ML^2T^{-3}I^{-1})^2[ε0​V2]=(M−1L−3T4I2)(ML2T−3I−1)2 =(M−1L−3T4I2)(M2L4T−6I−2)=MLT−2= (M^{-1}L^{-3}T^4I^2)(M^2L^4T^{-6}I^{-2}) = MLT^{-2}=(M−1L−3T4I2)(M2L4T−6I−2)=MLT−2

Both sides have same dimensions. So A is correct.

  1. Check option B

ε0I\varepsilon_0 Iε0​I Dimension: [ε0I]=M−1L−3T4I3[\varepsilon_0 I] = M^{-1}L^{-3}T^4I^3[ε0​I]=M−1L−3T4I3

μ0V\mu_0 Vμ0​V Dimension: [μ0V]=(MLT−2I−2)(ML2T−3I−1)=M2L3T−5I−3[\mu_0 V] = (MLT^{-2}I^{-2})(ML^2T^{-3}I^{-1}) = M^2L^3T^{-5}I^{-3}[μ0​V]=(MLT−2I−2)(ML2T−3I−1)=M2L3T−5I−3

These are not equal. So B is incorrect.

  1. Check option C

ε0cV\varepsilon_0 c Vε0​cV Dimension: [ε0cV]=(M−1L−3T4I2)(LT−1)(ML2T−3I−1)[\varepsilon_0 c V] = (M^{-1}L^{-3}T^4I^2)(LT^{-1})(ML^2T^{-3}I^{-1})[ε0​cV]=(M−1L−3T4I2)(LT−1)(ML2T−3I−1) =I= I=I

Thus, [ε0cV]=[I][\varepsilon_0 c V] = [I][ε0​cV]=[I] So C is correct.

  1. Check option D

μ0cI\mu_0 c Iμ0​cI Dimension: [μ0cI]=(MLT−2I−2)(LT−1)(I)=ML2T−3I−1[\mu_0 c I] = (MLT^{-2}I^{-2})(LT^{-1})(I) = ML^2T^{-3}I^{-1}[μ0​cI]=(MLT−2I−2)(LT−1)(I)=ML2T−3I−1

ε0V\varepsilon_0 Vε0​V Dimension: [ε0V]=(M−1L−3T4I2)(ML2T−3I−1)=L−1TI[\varepsilon_0 V] = (M^{-1}L^{-3}T^4I^2)(ML^2T^{-3}I^{-1}) = L^{-1}TI[ε0​V]=(M−1L−3T4I2)(ML2T−3I−1)=L−1TI

These are not equal. So D is incorrect.

  1. Final answer

The dimensionally correct equations are: A, C\boxed{A,\ C}A, C​

  1. Comparison with stored answer

Stored correct answer: A,CA, CA,C

This matches our derived answer.

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