Question

A plane electromagnetic wave of frequency \(25 \mathrm{MHz}\) travels in free space along the \(x\) direction. At a particular point in space and time \(E \rightarrow=6.3 \mathrm{j} \wedge \mathrm{Vm}^{-1}\) then \(\mathrm{B} \rightarrow\) at this point is
(A) \(2.1 \times 10^{-B}\) in \(T\)
(B) \(2.1 \times 10^{-B} \mathrm{kN} \mathrm{T}\)
(C) \(1.89 \times 10^{9} \mathrm{k} \wedge T\)
(D) \(2.52 \times 10^{-7} \mathrm{kn} \mathrm{T}\)

Answer 1

The correct option is (B) \(2.1 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\).
Explanation:
$$
\begin{aligned}
&f=25 \mathrm{MHz}=25 \times 10^{6} \mathrm{~Hz} \\
&E^{-t}=6.3 \mathrm{j} \wedge \mathrm{V} / \mathrm{m} \\
&E_{\mathrm{D}}=6.3
\end{aligned}
$$
We know \(\left(E_{a} / B_{a}\right)=C\)
\(\therefore B_{0}=\left(E_{0} / C\right)=\left[6.3 /\left(3 \times 10^{8}\right)\right]=2.1 \times 10^{-8} \mathrm{~T}\)
\(\therefore E^{-}\)is along \(y\) direction \& wave is along \(\times\) direction hence \(B^{-}\)has to
be perpendicular to both \(\times \& y\) axes.
\(E^{-} \times B^{-}\)should be along \(\times\)direction.
\(B^{-}\)should be along \(z\) direction
\(B^{-}=B_{0} k \wedge\)
\(B^{-}=2.1 \times 10^{-8} k \wedge T\)

Answer 2

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Answer 3

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