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प्रश्न
Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is v = 50.0 MHz.
- Determine B0, ω, k, and λ.
- Find expressions for E and B.
उत्तर
Electric field amplitude, E0 = 120 N/C
Frequency of source, v = 50.0 MHz = 50 × 106 Hz
Speed of light, c = 3 × 108 m/s
(a) Magnitude of magnetic field strength is given as:
`"B"_0 = "E"_0/"c"`
= `120/(3 xx 10^8)`
= 4 × 10−7 T
= 400 nT
Angular frequency of the source is given as:
ω = 2πv
= 2π × 50 × 106
= 3.14 × 108 rad/s
Propagation constant is given as:
`"k" = ω/"c"`
= `(3.14 xx 10^8)/(3 xx 10^8)`
= 1.05 rad/m
Wavelength of the wave is given as:
`lambda = "c"/"v"`
= `(3 xx 10^8)/(50 xx 10^6)`
= 6.0 m
(b) Suppose the wave is propagating in the positive x-direction. Then, the electric field vector will be in the positive y-direction and the magnetic field vector will be in the positive z-direction. This is because all three vectors are mutually perpendicular.
Equation of the electric field vector is given as:
`vec("E") = "E"_0 sin ("k""x" - ω"t")hat"j"`
= `120 sin [1.05"x" - 3.14 xx 10^8"t"]hat"j"`
Magnetic field vector is given as:
`vec("B") = "B"_0 sin ("kx" - ω"t")hat"k"`
`vec("B") = (4 xx 10^-7) sin [1.05"x" - 3.14 xx 10^8"t"]hat"k"`
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