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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