A plane electromagnetic wave of frequency 28 | vedclass

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Question

(A) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by $E = cB$, where $c$ is the speed

Vedclass · Accepted Answer

$3.1 	imes 10^{-8} \, T$ along the positive $z$-direction

Vedclass · Answer

(A) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by $E = cB$, where $c$ is the speed of light in free space $(c \approx 3 	imes 10^8 \, m/s)$.Given:$E = 9.3 \, V/m$ along the positive $y$-direction $(+\hat{j})$.Direction of propagation is along the positive $x$-direction $(+\hat{i})$.Using the relation $B = \frac{E}{c}$:$B = \frac{9.3}{3 	imes 10^8} = 3.1 	imes 10^{-8} \, T$.The direction of the magnetic field is determined by the cross product of the direction of propagation and the electric field. Since the wave propagates in the $x$-direction $(\hat{i})$ and the electric field is in the $y$-direction $(\hat{j})$, the magnetic field must be in the direction of $\hat{i} 	imes \hat{j} = \hat{k}$, which is the positive $z$-direction.Therefore, the magnetic field is $3.1 	imes 10^{-8} \, T$ along the positive $z$-direction.

$A$ plane electromagnetic wave of frequency $28 \, MHz$ travels in free space along the positive $x$-direction. At a particular point in space and time, the electric field is $9.3 \, V/m$ along the positive $y$-direction. The magnetic field (in $T$) at that point is:

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