A plane electromagnetic wave of frequency 25 | vedclass

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Question

(B) The magnitude of the magnetic field $B$ is related to the electric field $E$ by the relation $B = \frac{E}{c}$,where $c = 3 	imes 10^8 \; m/s$ is

Vedclass · Accepted Answer

$2.1 	imes 10^{-8} \hat{k} \; T$

Vedclass · Answer

(B) The magnitude of the magnetic field $B$ is related to the electric field $E$ by the relation $B = \frac{E}{c}$,where $c = 3 	imes 10^8 \; m/s$ is the speed of light in free space.Substituting the given values: $B = \frac{6.3 \; V/m}{3 	imes 10^8 \; m/s} = 2.1 	imes 10^{-8} \; T$.To find the direction,we know that the electromagnetic wave propagates in the direction of the vector $\vec{E} 	imes \vec{B}$.Given that the wave propagates along the $x$-direction $(\hat{i})$ and the electric field is along the $y$-direction $(\hat{j})$,we have $\hat{j} 	imes \hat{B} = \hat{i}$.Since $\hat{j} 	imes \hat{k} = \hat{i}$,the magnetic field must be in the $z$-direction $(\hat{k})$.Therefore,$\vec{B} = 2.1 	imes 10^{-8} \hat{k} \; T$.

$A$ plane electromagnetic wave of frequency $25 \; MHz$ travels in free space along the $x$-direction. At a particular point in space and time,$\vec{E} = 6.3 \hat{j} \; V/m$. What is $\vec{B}$ at this point?

Similar Questions

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$A$ plane electromagnetic wave of frequency $25 \text{ MHz}$ travels in free space along the $X$-direction. At a particular point in space and time, the electric field is $\vec{E} = 6.3 \hat{j} \text{ V m}^{-1}$. What is the magnetic field $\vec{B}$ at that point?

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