Question Number 100068 by M±th+et+s last updated on 24/Jun/20

Answered by JDamian last updated on 24/Jun/20

Commented by M±th+et+s last updated on 24/Jun/20

Answered by smridha last updated on 25/Jun/20
![Maxwell diff:eq^n said us ▽^2 E^→ =𝛍_0 ε_0 (∂^2 E^→ /∂t^2 ) and ▽^2 B^→ =𝛍_0 𝛆_0 (∂^2 B^→ /∂t^2 ) where E^→ =electric field B^→ =magnetic field 𝛍_0 =magnetic permiability in free space 𝛆_0 =permitivity of free space [this is besically 3d wave eq^n this two eq^(ns) describe light as a electromagnetic wave] now analysis the dimention of the product of μ_0 and 𝛆_0 .. dim[𝛆_0 𝛍_0 ]=[M^(−1) L^(−1) T^4 ].[ML^(−1) T^(−2) ] =[L^(−2) T^2 ] now dim[(1/( (√(𝛍_0 𝛆_0 ))))]=[L.T^(−1) ]=dim(v) where v=velocity but this not any odinary velocity this is very special, the speed of light in free space denoted by C.. so our eq^n looks like: ▽^2 E^→ =(1/C^2 )(∂^2 E^→ /∂t^2 ) and ▽^2 B^→ =(1/C^2 ).(∂^2 B^→ /∂t^2 ) now speed of light: C=(1/( (√(𝛍_0 𝛆_0 ))))=(1/( (√(4𝛑×10^(−7) ×8.854×10^(−12) )))) =2.99795638×10^8 ≈3×10^8 m.s^(−1) here I used the experimental value of 𝛍_0 and 𝛆_0 .](https://www.tinkutara.com/question/Q100083.png)
Commented by Rasheed.Sindhi last updated on 25/Jun/20

Commented by M±th+et+s last updated on 24/Jun/20

Commented by smridha last updated on 24/Jun/20

Commented by Rasheed.Sindhi last updated on 25/Jun/20

Commented by smridha last updated on 25/Jun/20
