Question

A plane electromagnetic waves travels in a medium of relative permeability $1.61$ and relative permittivity $6.44$. If magnitude of magnetic intensity is $4.5\times {10}^{-2}{\text{Am}}^{-1}$ at a point, what will be the approximate magnitude of electric field intensity at that point?
(Given: Permeability of free space ${\mu }_0=4\pi \times {10}^{-7}{\text{NA}}^{-2}$, speed of light in vacuum $c=3\times {10}^8{\text{ms}}^{-1}$)

• A. $8.48{\text{Vm}}^{-1}$
• B. $6.75\times {10}^6{\text{Vm}}^{-1}$
• C. $16.96{\text{Vm}}^{-1}$
• D. $2.25\times {10}^{-2}{\text{Vm}}^{-1}$

Answer 1

The correct option is A $8.48{\text{Vm}}^{-1}$

$H=4.5\times {10}^{-2}$
So $B={\mu }_0\mu H$
Thus $E=\frac{c}{n}B$ (where $n\Rightarrow$ refractive index)
So $E=\frac{3\times {10}^8\times 4\pi \times {10}^{-7}\times 1.61\times 4.5\times {10}^{-2}}{\sqrt{1.61\times 6.44}}$
$E=8.48$