An electromagnetic wave of frequency 45 text{ | vedclass

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(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

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

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(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

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