At a certain point and instant in a plane electromagnetic wave travelling in vacuum, the electric field has a magnitude of \( 20\,\mathrm{V\,m^{-1}} \). The energy density of the magnetic field at that point and instant is (Take \( \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.77\times10^{-9}\,\mathrm{J\,m^{-3}} \)
In an electromagnetic wave in vacuum the fields at any point satisfy \( B=E/c \), with \( c^{2}=1/\mu_0\varepsilon_0 \). The magnetic energy density then equals the electric energy density at every point and instant:
\( \begin{aligned} u_B &= \frac{B^{2}}{2\mu_0} \\ &= \frac{E^{2}}{2\mu_0c^{2}} \\ &= \frac{1}{2}\varepsilon_0E^{2} \end{aligned} \)
With \( E=20\,\mathrm{V\,m^{-1}} \) at the given point and instant:
\( \begin{aligned} u_B &= \frac{1}{2}\varepsilon_0E^{2} \\ &= \frac{1}{2}\times8.85\times10^{-12}\times(20)^{2}\,\mathrm{J\,m^{-3}} \\ &= 1.77\times10^{-9}\,\mathrm{J\,m^{-3}} \end{aligned} \)