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JEE Advanced 2021Paper 2
Exam
JEE Advanced
Year
2021
Paper
Paper 2
Subject
Physics
Chapter
Units and Dimensions

JEE Advanced 2021 · Paper 2 · Physics · Units and Dimensions

A physical quantity → S is defined as → S = ( → E × → B ) μ _ 0 ....

A physical quantity S→\overset{\to}{S} is defined as S→=(E→×B→)μ0\overset{\to}{S} = \dfrac{\left(\overset{\to}{E} \times \overset{\to}{B}\right)}{{\mu}_{0}}. where E→\overset{\to}{E} is electric field, B→\overset{\to}{B} is magnetic field and μ0{\mu}_{0} is the permeability of free space. The dimensions of S→\overset{\to}{S} are the same as the dimensions of which of the following quantity/quantities?

Options

  1. AEnergyCharge×Current\dfrac{Energy}{Charge \times Current}
  2. BForceLength×Time\dfrac{Force}{Length \times Time}
  3. CEnergyVolume\dfrac{Energy}{Volume}
  4. DPowerArea\dfrac{Power}{Area}

Correct answer

B. ForceLength×Time\dfrac{Force}{Length \times Time}

Step-by-step solution

Dimension of& ForceLength×Time=MLT−2LT=MT−3\dfrac{Force}{Length \times Time}= \dfrac{MLT- 2}{LT} = {MT}^{- 3}

Dimensions of PowerArea=Force×VelocityArea=MLT−2×LT−1L2=MT−3\dfrac{Power}{Area} = \dfrac{Force \times Velocity}{Area} = \dfrac{{MLT}^{- 2} \times {LT}^{- 1}}{L2} = {MT}^{- 3},~

Dimension of Electric field& \left(E\right), \left[E\right] = {MA}^{- 1}{LT}^{- 3}</p><p>DimensionsofMagneticfield (B);[B]=MA−1T−2</p><p>Dimensions of Magnetic field~\left(B\right); \left[B\right] = {MA}^{- 1}T^{- 2}

Dimension of magnetic permeability~\left({μ}_{0}\right); \left[{μ}_{0}\right] = {MA}^{- 2}T^{- 2}L</p><p>Therefore,~dimensions of&S,\left[S\right] = \left[\dfrac{EB}{\mu0}\right] = \dfrac{\left({MA}^{- 1}{LT}^{- 3}\right)\left({MA}^{- 1}T^{- 2}\right)}{{MA}^{- 2}T^{- 2}L} = {MT}^{- 3}

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