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  •   <img alt="\frac{E^{2}}{\mu _{0}}" src="https://learn.careers360.com/latex-image/?%5C

  \frac{E^{2}}{\mu _{0}} has the dimensions (E = electric field, \mu_{0} = permeability of free space)

Option: 1

 [M2L3T–2A2]       


Option: 2

[MLT–4]


Option: 3

[ML3T–2]


Option: 4

[M–1L2TA–2]


Answers (1)

best_answer

 

As we learned

Work, Potential Energy, Kinetic Energy, Torque -

Dimension- \dpi{100} ML^{2}T^{-2}

Unit  \dpi{100} N-m/Joule

And for  Permeability of free space (\mu_{o})-

Dimension-  M^{1}L^{1}T^{-2}A^{-2} 

Unit- \frac{newton}{ampere^{2}},\frac{henry}{metre}

    And for Permittivity of free space (\epsilon _0)-

Dimension-\ M^{-1}L^{-3}T^{4}A^{2}^{}

Unit- \dpi{100} C^2N^{-1}m^{-2}

Speed of light in a vacuum is given as  c=\frac{1}{\sqrt{\mu _0\varepsilon _0}}

So 

\left [ \frac{E^{2}}{\mu_{0} } \right ]=\left [ \frac{\epsilon _{0}E^{2}}{\epsilon _{0}\mu_{0}} \right ]=\left [ \frac{energy/volume}{\left ( 1/speed of light \right )^{2}} \right ]=\left [ \frac{energy(speed)^{2}}{volume} \right ]=\left [ \frac{ML^{2}T^{-2}L^{2}T^{-2} }{L^{3}}\right ]=\left [ MLT^{-4} \right ]

Posted by

SANGALDEEP SINGH

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