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प्रश्न
The 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.
उत्तर
Maximum value of a magnetic field, `B_0 = 200 "uT"`
The speed of an electromagnetic wave is c.
So, maximum value of electric field,
`E_0 = cB_0`
`E_0 = c xx B_0 = 200 xx 10^-6 xx 3 xx 10^8`
`E_0 = 6 xx 10^4 "NC"^-1`
(b) Average energy density of a magnetic field,
`U_(av) = 1/(2u_0) B_0^2 = (200 xx 10^-6)^2/(2 xx 4pi xx 10^-7)`
`U_(av) = (4 xx 10^-8)/(8pi xx 10^-7) = 1/(20pi)`
`U_(av) = 0.0159 ≈ 0.016 "J/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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