Electromagnetic WavesMedium
Question
The electric field in a plane electromagnetic wave is given by
$E_{z} = 60cos\left( 5x + 1.5 \times 10^{9}t \right)V/m$.
Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :
Options
A.$\ B_{y} = 2 \times 10^{- 7}cos\left( 5x + 1.5 \times 10^{9}t \right)T$
B.$\ B_{x} = 2 \times 10^{- 7}cos\left( 5x + 1.5 \times 10^{9}t \right)T$
C.$B_{z} = 60cos\left( 5x + 1.5 \times 10^{9}t \right)T$
D.$\ B_{y} = 60sin\left( 5x + 1.5 \times 10^{9}t \right)T$
Solution
In electromagnetic wave, $E$ and $B$ are in same phase and $B_{0} = \frac{E_{0}}{c}$; their planes are perpendicular to each other.
$$\begin{matrix} & \ \therefore B_{y} = \frac{60}{c}cos\left( 5x + 1.5 \times 10^{9}t \right)\top \\ & \ = \frac{60}{3 \times 10^{8}}cos\left( 5x + 1.5 \times 10^{9}t \right)\top \\ & B_{y} = 2 \times 10^{- 7}cos\left( 5x + 1.5 \times 10^{9}t \right)\top \end{matrix}$$
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