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8.Electromagnetic waves
medium
The magnetic field of a plane electromagnetic wave is given by
$\overrightarrow{ B }=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j } T$ The amplitude of the electric field would be.
A
$6 Vm ^{-1}$ along $x$-axis
B
$3 Vm ^{-1}$ along $z$-axis
C
$6 Vm ^{-1}$ along $z$-axis
D
$2 \times 10^{-8} Vm ^{-1}$ along $z$-axis
(JEE MAIN-2022)
Solution
$c =\frac{ E _{0}}{ B _{0}} \Rightarrow E _{0}= cB _{0}$
$E _{0}=\left(3 \times 10^{8}\right)\left(2 \times 10^{-8}\right)$
$E _{0}=6 Vm ^{-1}$
As, $\overrightarrow{ B }=$ along $y$-axis
$\overrightarrow{ v }$ =along negative $x$-axis
hence $\quad \overrightarrow{ E }_{0}=$ along $z$-axis
Standard 12
Physics
