A radiowave has a maximum electric field inte | vedclass

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

Question

(D) For an electromagnetic wave,the relationship between the maximum electric field intensity $(E_0)$ and the maximum magnetic flux density $(B_0)$ is

Vedclass · Accepted Answer

$3.3 	imes 10^{-13} \, T$

Vedclass · Answer

(D) For an electromagnetic wave,the relationship between the maximum electric field intensity $(E_0)$ and the maximum magnetic flux density $(B_0)$ is given by the equation: $B_0 = \frac{E_0}{c}$,where $c$ is the speed of light in vacuum $(c \approx 3 	imes 10^8 \, m/s)$.Given: $E_0 = 10^{-4} \, V/m$.Substituting the values: $B_0 = \frac{10^{-4}}{3 	imes 10^8} \, T$.$B_0 = \frac{1}{3} 	imes 10^{-12} \, T = 0.333 	imes 10^{-12} \, T = 3.33 	imes 10^{-13} \, T$.Therefore,the correct option is $D$.

$A$ radiowave has a maximum electric field intensity of $10^{-4} \, V/m$ on arrival at a receiving antenna. The maximum magnetic flux density of such a wave is:

Similar Questions

Which of the following statements is incorrect?

Let $E_0$ and $B_0$ denote the amplitude of the electric and magnetic fields of a plane electromagnetic wave in air. The magnitude of the average momentum transferred per unit area and per unit time to a totally absorbing surface is

$A$ plane electromagnetic $(EM)$ wave is propagating along the $x$-direction. It has a wavelength of $4 \text{ mm}$. If the electric field is in the $y$-direction with a maximum magnitude of $60 \text{ Vm}^{-1}$,the equation for the magnetic field is:

$A$ plane electromagnetic wave is incident on a material surface. If the wave delivers momentum $p$ and energy $E$,then

$A$ plane electromagnetic wave propagating along the y-direction can have the following pair of electric field $(\vec{E})$ and magnetic field $(\vec{B})$ components.

Vedclass Test Series

Exam Paper Generator

Online Exam Module