An electromagnetic wave of frequency 45 text{ | vedclass

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

Question

(C) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by $B = \frac{E}{c}$.Given, $E = 75

Vedclass · Accepted Answer

$2.5 	imes 10^{-6} \hat{k} 	ext{ T}$

Vedclass · Answer

(C) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by $B = \frac{E}{c}$.Given, $E = 750 	ext{ NC}^{-1}$.The speed of light in free space is $c = 3 	imes 10^8 	ext{ ms}^{-1}$.Substituting these values, we get:$B = \frac{750}{3 	imes 10^8} = 250 	imes 10^{-8} = 2.5 	imes 10^{-6} 	ext{ T}$.The electromagnetic wave travels along the $X$-axis ($\hat{i}$ direction) and the electric field is along the $Y$-axis ($\hat{j}$ direction).Since the direction of propagation is given by the cross product of the electric and magnetic fields $(\vec{E} 	imes \vec{B})$, the direction of the magnetic field must be along the $Z$-axis ($\hat{k}$ direction) because $\hat{j} 	imes \hat{k} = \hat{i}$.Therefore, the magnetic field is $2.5 	imes 10^{-6} \hat{k} 	ext{ T}$.

An electromagnetic wave of frequency $45 \text{ MHz}$ travels in free space along the $X$-axis. At some point and at some instant, the electric field has a maximum value of $750 \text{ NC}^{-1}$ along the $Y$-axis. The magnetic field at this position and time is

Similar Questions

Find the direction of vibration of the electric field if the vibration of the magnetic field is in the positive $x-$ axis and the propagation of the electromagnetic wave is along the positive $y-$ axis.

The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0 = 510 \; nT$. What is the amplitude of the electric field (in $N/C$) part of the wave?

The electric field in a plane electromagnetic wave is given by $E_z = 60 \cos(5x + 1.5 \times 10^9 t) \text{ V/m}$. Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

The velocity of an electromagnetic wave is in the direction of

The intensity of light from a source is $\left( \frac{500}{\pi} \right) \, W/m^2$. Find the amplitude of the electric field in this wave.

Vedclass Test Series

Exam Paper Generator

Online Exam Module