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         \[{{E}_{0}}=45V/m\]