The magnetic field of a plane electromagnetic | vedclass

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

Question

(D) The force on a stationary charge $Q$ in an electromagnetic wave is due to the electric field $\vec E$ only,as $\vec F = Q\vec E$. For an electroma

Vedclass · Accepted Answer

$0.6 \, N$

Vedclass · Answer

(D) The force on a stationary charge $Q$ in an electromagnetic wave is due to the electric field $\vec E$ only,as $\vec F = Q\vec E$. For an electromagnetic wave,$\vec E = c(\vec B 	imes \hat k)$,where $\hat k$ is the direction of propagation ($z$-axis). Given $\vec B = B_0 \cos(kz - \omega t) \hat i + B_1 \cos(kz - \omega t) \hat j$. Then $\vec E = c [ (B_0 \hat i + B_1 \hat j) 	imes \hat k ] \cos(kz - \omega t) = c [ -B_0 \hat j + B_1 \hat i ] \cos(kz - \omega t)$. The magnitude of the electric field is $E = c \sqrt{B_0^2 + B_1^2} |\cos(kz - \omega t)|$. The maximum force is $F_0 = Q E_{max} = Q c \sqrt{B_0^2 + B_1^2}$. Substituting values: $F_0 = 10^{-4} 	imes 3 	imes 10^8 	imes \sqrt{(3 	imes 10^{-5})^2 + (2 	imes 10^{-6})^2} \approx 10^{-4} 	imes 3 	imes 10^8 	imes 3 	imes 10^{-5} = 0.9 \, N$. The rms force is $F_{rms} = \frac{F_0}{\sqrt{2}} = \frac{0.9}{1.414} \approx 0.636 \, N$. Thus,the closest value is $0.6 \, N$.

Similar Questions

The electric and the magnetic field,associated with an electromagnetic wave propagating along the $+z$-axis,can be represented by

When an electromagnetic wave enters the ionized layer of the ionosphere,the relative permittivity of the ionized layer is found to be:

If the magnetic field in a plane electromagnetic wave is given by $\overrightarrow{B} = 3 \times 10^{-8} \sin (1.6 \times 10^{3} x + 48 \times 10^{10} t) \hat{j} \; T$,then what will be the expression for the electric field?

$A$ radio receiver antenna that is $4 \, m$ long is oriented along the direction of the electromagnetic wave and it receives a signal of intensity $8 \times 10^{-16} \, W/m^2$. The maximum instantaneous potential difference across the two ends of the antenna is (in $ \, \mu V$)

Electromagnetic waves are incident normally on a perfectly reflecting surface having surface area $A$. If $I$ is the intensity of the incident electromagnetic radiation and $c$ is the speed of light in vacuum,the force exerted by the electromagnetic wave on the reflecting surface is

Vedclass Test Series

Exam Paper Generator

Online Exam Module