JEE Challenger
More from Units and Measurements
PhyMedAdv 2021 • Paper 2

Dimensional Analysis of Poynting Vector Analogue

A physical quantity S⃗\vec{S} is defined as S⃗=(E⃗×B⃗)/μ0\vec{S} = (\vec{E} \times \vec{B})/\mu_0, where E⃗\vec{E} is the electric field, B⃗\vec{B} is the magnetic field, and μ0\mu_0 denotes the magnetic permeability of free space. The dimensions of S⃗\vec{S} match the dimensions of which of the following physical quantity(ies)?

Options

A

EnergyCharge×Current\frac{\text{Energy}}{\text{Charge} \times \text{Current}}

B

ForceLength×Time\frac{\text{Force}}{\text{Length} \times \text{Time}}

Correct
C

EnergyVolume\frac{\text{Energy}}{\text{Volume}}

D

PowerArea\frac{\text{Power}}{\text{Area}}

Correct

Topics & Concepts

Units and MeasurementsElectromagnetic Waves

Step-by-Step Solution

To determine which of the given options match the dimensions of the physical quantity S⃗\vec{S}, we first find the dimensional formula of S⃗=E⃗×B⃗μ0\vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0}, which is known as the Poynting vector.

Step 1: Dimensional Analysis of S⃗\vec{S}

  1. Electric Field (E⃗\vec{E}): [E⃗]=[Force][Charge]=M L T−2A T=M L T−3A−1[\vec{E}] = \frac{[\text{Force}]}{[\text{Charge}]} = \frac{\text{M L T}^{-2}}{\text{A T}} = \text{M L T}^{-3}\text{A}^{-1}

  2. Magnetic Field (B⃗\vec{B}): From the magnetic force equation F=qvBF = q v B: [B⃗]=[Force][Charge]×[Velocity]=M L T−2(A T)(L T−1)=M T−2A−1[\vec{B}] = \frac{[\text{Force}]}{[\text{Charge}] \times [\text{Velocity}]} = \frac{\text{M L T}^{-2}}{(\text{A T})(\text{L T}^{-1})} = \text{M T}^{-2}\text{A}^{-1}

  3. Magnetic Permeability (μ0\mu_0): From Ampere's law / force per unit length between two parallel conductors FL=μ0I22πd\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}: [μ0]=[Force][Current]2=M L T−2A2=M L T−2A−2[\mu_0] = \frac{[\text{Force}]}{[\text{Current}]^2} = \frac{\text{M L T}^{-2}}{\text{A}^2} = \text{M L T}^{-2}\text{A}^{-2}

  4. Dimensions of S⃗\vec{S}: [S⃗]=[E⃗][B⃗][μ0]=(M L T−3A−1)(M T−2A−1)M L T−2A−2=M2L T−5A−2M L T−2A−2=M T−3[\vec{S}] = \frac{[\vec{E}][\vec{B}]}{[\mu_0]} = \frac{(\text{M L T}^{-3}\text{A}^{-1})(\text{M T}^{-2}\text{A}^{-1})}{\text{M L T}^{-2}\text{A}^{-2}} = \frac{\text{M}^2\text{L T}^{-5}\text{A}^{-2}}{\text{M L T}^{-2}\text{A}^{-2}} = \text{M T}^{-3}

Step 2: Dimensional Analysis of the Given Options

  • Option (A): [EnergyCharge×Current]=M L2T−2(A T)(A)=M L2T−3A−2≠[S⃗]\left[\frac{\text{Energy}}{\text{Charge} \times \text{Current}}\right] = \frac{\text{M L}^2\text{T}^{-2}}{(\text{A T})(\text{A})} = \text{M L}^2\text{T}^{-3}\text{A}^{-2} \neq [\vec{S}]

  • Option (B): [ForceLength×Time]=M L T−2L⋅T=M T−3=[S⃗]\left[\frac{\text{Force}}{\text{Length} \times \text{Time}}\right] = \frac{\text{M L T}^{-2}}{\text{L} \cdot \text{T}} = \text{M T}^{-3} = [\vec{S}]

  • Option (C): [EnergyVolume]=M L2T−2L3=M L−1T−2≠[S⃗]\left[\frac{\text{Energy}}{\text{Volume}}\right] = \frac{\text{M L}^2\text{T}^{-2}}{\text{L}^3} = \text{M L}^{-1}\text{T}^{-2} \neq [\vec{S}]

  • Option (D): [PowerArea]=M L2T−3L2=M T−3=[S⃗]\left[\frac{\text{Power}}{\text{Area}}\right] = \frac{\text{M L}^2\text{T}^{-3}}{\text{L}^2} = \text{M T}^{-3} = [\vec{S}]

Conclusion

The physical quantities given in options (B) and (D) have the same dimensions as S⃗\vec{S}.

Correct Options: (B) and (D)

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