Question 17: EM waves

Question

A cube of unit volume contains $35 	imes 10^7$ photons of frequency $10^{15} \mathrm{~Hz}$. If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is $\alpha 	imes 10^{-9} \mathrm{~T}$. Taking permeability of free space $\mu_0=4 \pi 	imes 10^{-7} \mathrm{Tm} / \mathrm{A}$, Planck's constant $h=6 	imes 10^{-34} \mathrm{Js}$ and $\pi=\frac{22}{7}$, the value of $\alpha$ is ____________.

Accepted Answer

As the cube has unit volume so energy density, $u=\frac{E}{1}=E$
where, $E=$ total energy
Energy of a photon, $E_1=h 
u$
So for n photon, $E=n h 
u$
Here, the question mentions "average energy contained in the electromagnetic waves within the same volume" which means we have to consider standing wave in a cavity (since the photons are in a cube).
Average energy density for a standing wave : $\langle uangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 E_0{ }^2+\frac{1}{2} \frac{B_0{ }^2}{\mu_0}ight)$
The factor $\frac{1}{2}$ comes from averaging $\sin ^2(k x)$
over the volume. $\left\langle\sin ^2(k x)ightangle=\frac{1}{2}$
For an $E M$ wave $E=C B$
$$ 	ext { so } E_0=C B_0 $$
Hence, $\langle uangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0\left(c B_0ight)^2+\frac{B_0^2}{2 \mu_0}ight)$
$$ =\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 c^2 B_0^2+\frac{B_0^2}{2 \mu_0}ight) $$
$$ \begin{aligned} \Rightarrow\langle uangle=\frac{1}{2} & {\left[\frac{1}{2}\left(\frac{1}{\mu_0 c^2}ight) c^2 B_0^2+\frac{B_0^2}{2 \mu_0}ight]\left(\begin{array}{l} As, c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \ 	ext { so } \varepsilon_0=\frac{1}{\mu_0 c^2} \end{array}ight.} \ \Rightarrow\langle uangle & =\frac{1}{2}\left[\frac{B_0^2}{2 \mu_0}+\frac{B_0^2}{2 \mu_0}ight] \ & \Rightarrow\langle uangle=\frac{B_0^2}{2 \mu_0} \end{aligned} $$
Given that this is equal to the photon energy density. so $\frac{B_0^2}{2 \mu_0}=n h 
u$
Total energy of photons.
$$ \begin{aligned} E & =n h 
u \ & =35 	imes 10^7 	imes 6 	imes 10^{-34} 	imes 1.15 \ \Rightarrow E & =210 	imes 10^{-12} \mathrm{~J} \end{aligned} $$
For $E M$ waves : $E=\frac{B_0{ }^2}{2 \mu_0}$
So $\frac{B_0^2}{2 \mu_0}=210 	imes 10^{-12}$
$$ \begin{aligned} & \Rightarrow B_0^2=2 	imes 4 	imes \frac{22}{7} 	imes 10^{-7} 	imes 210 	imes 10^{-12} \ & \Rightarrow B_0^2=5280 	imes 10^{-19} \ & \Rightarrow B_0^2=528 	imes 10^{-18} \ & \Rightarrow B_0=\sqrt{528} 	imes 10^{-9} \ & \Rightarrow B_0=22.978 	imes 10^{-9} \mathrm{~T} \ & \ & 	ext { = } \alpha 	imes 10^{-9} \mathrm{~T} 	ext { (given) } \ &So, \alpha=22.98 	ext { Ans. } \end{aligned} $$

EM waves

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