An electromagnetic wave going through vacuum is described by E=E_… - Vidyadip

MCQ

An electromagnetic wave going through vacuum is described by $\text{E}=\text{E}_0\sin(\text{kx}-\omega\text{t}),\text{ B}=\text{B}_0\sin(\text{kx}-\omega\text{t})$ Then:

✓
$\text{E}_0\text{k}=\text{B}_0\omega$
B
$\text{E}_0\text{B}_0=\omega\text{k}$
C
$\text{E}_0\omega=\text{B}_0\text{k}$
D
None of these.

✓

Answer

Correct option: A.

$\text{E}_0\text{k}=\text{B}_0\omega$

The relation between $E _0$ and $B _0$ id given by $\frac{\text{E}_0}{\text{B}_0}=\text{c}\ ....(\text{i})$
Here, $c =$ Speed of the electromagnetic wave,
The relation between $\omega ($the angular frequency$)$ and $k\ ($wave number$),$
$\frac{\omega}{\text{k}}=\text{c}\ ...(\text{ii})$
Therefore, from $(i)$ and $(ii),$ we get
$\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}=\text{c}$
$\text{E}_0\text{k}=\text{B}_0\omega$

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