In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $$2.0 \times 10^{10} \mathrm{~Hz}$$ and amplitude $$48 ~\mathrm{Vm}^{-1}$$. Then the amplitude of oscillating magnetic field is : (Speed of light in free space $$=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$$ )
The magnetic field of a plane electromagnetic wave is given by $$\overrightarrow B = 3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat j$$, then the associated electric field will be :
The ratio of the magnitude of the magnetic field and electric field intensity of a plane electromagnetic wave in free space of permeability $${\mu _0}$$ and permittivity $${\varepsilon _0}$$ is (Given that c - velocity) of light in free space
Match List - I with List - II
| List - I (Electromagnetic waves) |
List - II (Wavelength) |
||
|---|---|---|---|
| (a) | AM radio waves | (i) | $${10^{ - 10}}$$ m |
| (b) | Microwaves | (ii) | $${10^2}$$ m |
| (c) | Infrared radiations | (iii) | $${10^{ - 2}}$$ m |
| (d) | X-rays | (iv) | $${10^{ - 4}}$$ m |
Choose the correct answer from the options given below