• # question_answer The magnetic field in an electromagnetic wave, oscillates with an amplitude of $2\times {{10}^{-7}}$ T. What would be the amplitude of the electric field, if the wave is travelling in water of refractive index 4/3?

Answer:

Let, ${{v}_{w}}$ be the speed of electromagnetic waves in water. As we know, ${{\mu }_{w}}=C/{{v}_{w}}$ where, $\mu$ = refractive index of water and       c = speed of light in air $\Rightarrow$   ${{v}_{w}}=\frac{c}{{{\mu }_{w}}}=\frac{3\times {{10}^{8}}}{4/3}$ $[\because given\,{{\mu }_{w}}=4/3]$             $=2.25\times {{10}^{8}}m{{s}^{-1}}$ Given,   ${{B}_{0}}=2\times {{10}^{-7}}T$ We know that, ${{v}_{w}}={{E}_{0}}/{{B}_{0}}$ $\Rightarrow$   ${{E}_{0}}{{B}_{0}}{{v}_{w}}=2\times {{10}^{-7}}\times 2.25\times {{10}^{8}}$ or         ${{E}_{0}}=45V/m$

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