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In a plane electomagnetic wave, the elec...

In a plane electomagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx106(10)Hz` and amplitude `48Vm^(-1)`.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the ocillating magnetic field ?
(c ) Show that the average energy density of the ` vecE` field equals the average energy density of the `vecB` field.

Text Solution

Verified by Experts

(a) `lambda=(c )/(v)=(3xx10^(8))/(2.0xx10^(10))=1.5xx10^(-2)m`
(b) `B_(0)=(E_(0))/(c )=(48Vm^(-1))/(3xx10^(8)ms^(-1))=1.6xx10^(-7)T`
(c ) We know that average energy density of `vecE` field, `u_(E)=(1)/(2)epsilon_(0)E_("rms")^(2)` and average energy density of `vecB` field, `u_(B)=(1)/(2mu_(0))B_("rms")^(2)`.
Now `E_("rms")=cB_("rms") and c=(1)/(sqrt(mu_(0)E_(0)))`
`therefore" "u_(E)=(1)/(2)epsilon_(0)E_("rms")^(2)=(1)/(2)epsilon_(0)c^(2)B_("rms")^(2)=(1)/(2)epsilon_(0)((1)/(sqrt(mu_(0)epsilon_(0))))^(2)B_("rms")^(2)=(1)/(2mu_(0))B_("rms")^(2)=u_(B)`
Thus, it is clear the average energy densities of `vecE` field and `vecB` field are equal.
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