Question
Derive the equation of energy density for electromagnetic wave.
✓
Answer
$\rightarrow $Energy density : "Energy stored per unit volume is called Energy density." An electromagnetic wave contains Electric field and magnetic field both.
$\rightarrow$ Energy density associated with Electric field, $\varrho_{ E }=\frac{1}{2} \varepsilon_0 E ^2$
$\rightarrow$ Energy$-$density associated with magnetic field, $\varrho _{ B }=\frac{ B ^2}{2 \mu_0}$
$\rightarrow$ Total Energy$-$density associated with $EM$ wave,
$\varrho =\varrho_{ E }+\varrho_{B}$
$\therefore =\frac{1}{2} \varepsilon_0 E ^2+\frac{ B ^2}{2 \mu_0}$
$\rightarrow$ But the magnitudes of electric field and magnetic field change as per sine or cosine functions in an $EM$ wave.
Hence, in the equation, the rms value of electric field and magnetic field is considered.
$\rightarrow$ Energy$-$density $\varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{ B _{\text {rms }}^2}{2 \mu_0}$
But $c ^2=\frac{1}{\mu_0 \varepsilon_0} \therefore \mu_0=\frac{1}{ c ^2 \varepsilon_0}$
and $\frac{E_{r m s}}{B_{r m s}}=c \therefore B_{r m s}=\frac{E_{r m s}}{c}$
$\rightarrow$ Substituting both these values in eq. $(1),$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{\frac{ E _{\text {rms }}^2}{ c ^2}}{2\left(\frac{1}{ c ^2 \varepsilon_0}\right)}$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2$
$\therefore \varrho=\varepsilon_0 E _{\text {rms }}^2$
$\rightarrow$ In similar manner, $\varrho=\frac{B_{\text {rms }}^2}{\mu_0}$ can also be derived.
$\rightarrow$ Energy density associated with Electric field, $\varrho_{ E }=\frac{1}{2} \varepsilon_0 E ^2$
$\rightarrow$ Energy$-$density associated with magnetic field, $\varrho _{ B }=\frac{ B ^2}{2 \mu_0}$
$\rightarrow$ Total Energy$-$density associated with $EM$ wave,
$\varrho =\varrho_{ E }+\varrho_{B}$
$\therefore =\frac{1}{2} \varepsilon_0 E ^2+\frac{ B ^2}{2 \mu_0}$
$\rightarrow$ But the magnitudes of electric field and magnetic field change as per sine or cosine functions in an $EM$ wave.
Hence, in the equation, the rms value of electric field and magnetic field is considered.
$\rightarrow$ Energy$-$density $\varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{ B _{\text {rms }}^2}{2 \mu_0}$
But $c ^2=\frac{1}{\mu_0 \varepsilon_0} \therefore \mu_0=\frac{1}{ c ^2 \varepsilon_0}$
and $\frac{E_{r m s}}{B_{r m s}}=c \therefore B_{r m s}=\frac{E_{r m s}}{c}$
$\rightarrow$ Substituting both these values in eq. $(1),$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{\frac{ E _{\text {rms }}^2}{ c ^2}}{2\left(\frac{1}{ c ^2 \varepsilon_0}\right)}$
$\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2$
$\therefore \varrho=\varepsilon_0 E _{\text {rms }}^2$
$\rightarrow$ In similar manner, $\varrho=\frac{B_{\text {rms }}^2}{\mu_0}$ can also be derived.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The nucleus $^{235}_{92}\text{Y},$ initially at rest, decays into $^{231}_{90}\text{X}$ by emitting an $\alpha-$particle
$^{235}_{92}\text{Y}\xrightarrow{ \ \ \ \ \ \ \ \ } \ ^{231}_{90}\text{X}+ \ ^4_2\text{He}+\text{energy}.$
The binding energies per nucleon of the parent nucleus, the daughter nucleus and $\alpha-$particle are $7.8\ MeV, 7.835\ MeV$ and $7.07\ MeV,$ respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted $\alpha-$particle. $($Mass of $\alpha-$ particle $= 6.68 \times 10^{–27}kg).$
→$^{235}_{92}\text{Y}\xrightarrow{ \ \ \ \ \ \ \ \ } \ ^{231}_{90}\text{X}+ \ ^4_2\text{He}+\text{energy}.$
The binding energies per nucleon of the parent nucleus, the daughter nucleus and $\alpha-$particle are $7.8\ MeV, 7.835\ MeV$ and $7.07\ MeV,$ respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted $\alpha-$particle. $($Mass of $\alpha-$ particle $= 6.68 \times 10^{–27}kg).$
A certain material has refractive indices 1.56, 1.60 and 1.68 for red, yellow and violet light respectively.
- Calculate the dispersive power.
- Find the angular dispersion produced by a thin prism of angle 6° made of this material.
The intensity of sound from a point source is $1.0 \times 10^{-6}W/m^2$ at a distance of $5.0m$ from the source. What will be the intensity at a distance of $25m$ from the source?
→Let x and a stand for distance. Is $\int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\frac{1}{\text{a}}\sin^{-1}\frac{\text{a}}{\text{x}}$ dimensionally correct?
→A circular ring of radius r made of a nonconducting material is placed with its axis parallel to a uniform electric field. The ring is rotated about a diameter through $180^\circ$ . Does the flux of electric field change? If yes, does it decrease or increase?
→State the limitations of ohm's law.
- In the following diagram, which bulb out of $B_1$ and $B_2$ will glow and why?
- Draw a diagram of an illuminated $p-n$ junction solar cell.
- Explain briefly the three processes due to which generation of emf takes place in a solar cell.
In the figure a long uniform potentiometer wire $AB$ is having a constant potential gradient along its length. The null points for the two primary cells of emfs $\varepsilon_{1}$ and $\varepsilon_{2}$
connected in the manner shown are obtained at a distance of $120 \ cm $ and $300 \ cm$ from the end $A$.
Find $(i) \varepsilon_{1} / \varepsilon_{2}$ and $(ii)$ position of null point for the cell $\varepsilon_{1}$. How is the sensitivity of a potentiometer increased?
Using Kirchoffs rules determine the value of unknown resistance $R$ in the circuit so that no current flows through 4 $\Omega$ resistance. Also find the potential difference between $A$ and $D.$
→connected in the manner shown are obtained at a distance of $120 \ cm $ and $300 \ cm$ from the end $A$.
Find $(i) \varepsilon_{1} / \varepsilon_{2}$ and $(ii)$ position of null point for the cell $\varepsilon_{1}$. How is the sensitivity of a potentiometer increased?
Using Kirchoffs rules determine the value of unknown resistance $R$ in the circuit so that no current flows through 4 $\Omega$ resistance. Also find the potential difference between $A$ and $D.$
Using Gauss's law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.
A sphere of mass $20\ kg$ is suspended by a metal wire of unstretched length $4m$ and diameter $1\ mm$. When in equilibrium, there is a clear gap of $2\ mm$ between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle $\theta$ with the vertical and is released. Find the maximum value of $\theta$ so that the sphere does not rub the floor. Young's modulus of the metal of the wire is $2.0 \times 10^{11}\ N/m^2$. Make appropriate approximations.
→