The electric field of a plane electromagnetic | vedclass

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

Question

(B) The amplitude of the electric field $(E_0)$ and magnetic field $(B_0)$ are related by the equation $E_0 / B_0 = c$,where $c$ is the speed of light

Vedclass · Accepted Answer

$8.86 	imes 10^{-12}$

Vedclass · Answer

(B) The amplitude of the electric field $(E_0)$ and magnetic field $(B_0)$ are related by the equation $E_0 / B_0 = c$,where $c$ is the speed of light.The average energy density of the magnetic field is given by $u_B = \frac{1}{4} \frac{B_0^2}{\mu_0}$.Substituting $B_0 = E_0 / c$,we get $u_B = \frac{1}{4} \frac{E_0^2}{\mu_0 c^2}$.Since $c^2 = 1 / (\mu_0 \varepsilon_0)$,we can rewrite the expression as $u_B = \frac{1}{4} \varepsilon_0 E_0^2$.Given $E_0 = 2 \, Vm^{-1}$ and $\varepsilon_0 = 8.854 	imes 10^{-12} \, F m^{-1}$,we have:$u_B = \frac{1}{4} 	imes 8.854 	imes 10^{-12} 	imes (2)^2$$u_B = 8.854 	imes 10^{-12} \approx 8.86 	imes 10^{-12} \, J m^{-3}$.

The electric field of a plane electromagnetic wave varies with time with an amplitude of $2 \, Vm^{-1}$ propagating along the $z$-axis. The average energy density of the magnetic field (in $J \, m^{-3}$) is:

Similar Questions

If $3 \cos x \neq 2 \sin x$,then the general solution of $\sin^2 x - \cos 2x = 2 - \sin 2x$ is

The solubility of $N_2$ in water at $300 \ K$ and $500 \ torr$ partial pressure is $0.01 \ g \ L^{-1}$. The solubility (in $g \ L^{-1}$) at $750 \ torr$ partial pressure is

The equations of motion of a projectile are given by $x = 36 t$ metre and $2 y = 96 t - 9.8 t^2$ metre. The angle of projection is:

$A$ green light is incident from the water to the air-water interface at the critical angle $(\theta_c)$. Select the correct statement.

In any fission process,the ratio $\frac{\text{mass of fission products}}{\text{mass of parent nucleus}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module