NEET-XII-Physics
40: Electromagnetic Waves
- #7The magnetic field in a plane electromagnetic wave is given by
B = (200 µT) sin [(4.0 × 1015s-1)(t-x/c)].
Find the maximum electric field and the average energy density corresponding to the electric field.Ans : Maximum value of a magnetic field, B0 = 200 `` \,\mathrm{\,\mu \,}``T
The speed of an electromagnetic wave is c.
So, maximum value of electric field,
`` {E}_{0}=c{B}_{0}``
`` {E}_{0}=\,\mathrm{\,c\,}\times {B}_{0}=200\times {10}^{-6}\times 3\times {10}^{8}``
`` {E}_{0}=6\times {10}^{4}{\,\mathrm{\,NC\,}}^{-1}``
(b) Average energy density of a magnetic field,
`` {U}_{av}=\frac{1}{2{\,\mathrm{\,\mu \,}}_{0}}{{B}_{0}}^{2}=\frac{(200\times {10}^{-6}{)}^{2}}{2\times 4\,\mathrm{\,\pi \,}\times {10}^{-7}}``
`` {U}_{av}=\frac{4\times {10}^{-8}}{8\,\mathrm{\,\pi \,}\times {10}^{-7}}=\frac{1}{20\,\mathrm{\,\pi \,}}``
`` {U}_{av}=0.0159\approx 0.016\,\mathrm{\,J\,}/{\,\mathrm{\,m\,}}^{3}``
For an electromagnetic wave, energy is shared equally between the electric and magnetic fields.
Hence, energy density of the electric field will be equal to the energy density of the magnetic field.
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