NEET-XII-Physics
40: Electromagnetic Waves
- #5Using B = µ0 H, find the ratio E0/H0 for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.Ans : Given, B = µ0H
For vacuum we can rewrite this equation as,
B0 = µ0H0 ...(i)
Relation between magnetic field and electric field for vacuum is given as,
B0 = µ0`` {\in }_{0}``cE0 ...(ii)
From equation (i) by (ii),
`` {\,\mathrm{\,\mu \,}}_{0}{H}_{0}={\,\mathrm{\,\mu \,}}_{0}{\in }_{0}c{E}_{0}``
`` \Rightarrow \frac{{E}_{0}}{{H}_{0}}=\frac{1}{{\in }_{0}c}``
`` \Rightarrow \frac{{E}_{0}}{{H}_{0}}=\frac{1}{8.85\times {10}^{-12}\times 3\times {10}^{8}}``
`` \Rightarrow \frac{{E}_{0}}{{H}_{0}}\approx 377\,\mathrm{\,\Omega \,}``
`` \,\mathrm{\,Dimension\,}\text{of}\frac{1}{{\in }_{0}c}=\frac{1}{\left[{\,\mathrm{\,LT\,}}^{-1}\right]\left[{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-3}{\,\mathrm{\,T\,}}^{4}{\,\mathrm{\,A\,}}^{2}\right]}``
`` =\frac{1}{{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-2}{\,\mathrm{\,T\,}}^{3}{\,\mathrm{\,A\,}}^{2}}``
`` ={\,\mathrm{\,M\,}}^{4}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-3}{\,\mathrm{\,A\,}}^{-2}=\left[R\right]``
Page No 339: