Due to the presence of an em-wave whose elect | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The energy density of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E^2$. The total energy $U$ contained in a cylinder of length

Vedclass · Accepted Answer

$200 \sin(\omega t - kx) \ NC^{-1}$

Vedclass · Answer

(B) The energy density of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E^2$. The total energy $U$ contained in a cylinder of length $L$ and radius $R$ is $U = u 	imes V = \frac{1}{2} \epsilon_0 E^2 	imes (\pi R^2 L)$. Given that the energy $U$ remains the same for both cylinders,we have $U_1 = U_2$. $\frac{1}{2} \epsilon_0 E_1^2 \pi R_1^2 L_1 = \frac{1}{2} \epsilon_0 E_2^2 \pi R_2^2 L_2$. Since $L_1 = L_2$ and $R_2 = \frac{R_1}{2}$,the equation simplifies to $E_1^2 R_1^2 = E_2^2 (\frac{R_1}{2})^2$. $E_1^2 R_1^2 = E_2^2 \frac{R_1^2}{4}$. $E_2^2 = 4 E_1^2$,which implies $E_2 = 2 E_1$. Given $E_1 = 100 \ NC^{-1}$,we get $E_2 = 2 	imes 100 = 200 \ NC^{-1}$. Thus,the electric field is $200 \sin(\omega t - kx) \ NC^{-1}$.

Similar Questions

Electromagnetic waves are:

If $\overrightarrow{E}$ and $\overrightarrow{B}$ represent electric and magnetic field vectors of an electromagnetic wave,the direction of propagation of the electromagnetic wave is along . . . . . . .

The dielectric constant of a medium is $8$ and its relative permeability is $200$. If an electromagnetic wave of frequency $100 \text{ MHz}$ travels in this medium,then its wavelength is

$A$ plane electromagnetic wave of frequency $50\, MHz$ travels in free space along the positive $x-$ direction. At a particular point in space and time, $\vec E = 6.3\,\hat j\,V/m$. The corresponding magnetic field $\vec B$ at that point will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module