Modern Physics Ques 318
- A plane electromagnetic wave travels in free space along the $x$-direction. The electric field component of the wave at a particular point of space and time is $E=6 Vm^{-1}$ along $y$-direction. Its corresponding magnetic field component, $B$ would be
(Main 2019, 8 April I)
(a) $2 \times 10^{-8} T$ along $z$-direction
(b) $6 \times 10^{-8} T$ along $x$-direction
(c) $6 \times 10^{-8} T$ along $z$-direction
(d) $2 \times 10^{-8} T$ along $y$-direction
Show Answer
Solution:
Formula:
Relation Between The Magnetic Field Vector And The Electric Field Vector:
- Key Idea For an electromagnetic wave, ratio of magnitudes of electric and magnetic field is
$$ \frac{E}{B}=c $$
where, $c$ is the speed of electromagnetic wave in vacuum.
Given, $E=6 V / m, c=3 \times 10^{8} ms^{-1}$
So, $\quad B=\frac{E}{c}=\frac{6}{3 \times 10^{8}}=2 \times 10^{-8} T$
Also, direction of propagation of electromagnetic wave is given by
$$ \hat{\mathbf{n}}=\mathbf{E} \times \mathbf{B} $$
Here, $\hat{\mathbf{n}}=\hat{\mathbf{i}}$ and $\mathbf{E}=$ Unit vector of electric field $(\hat{\mathbf{j}})$
$\mathbf{B}=$ unit vector of magnetic field.
$\Rightarrow \quad \hat{\mathbf{i}}=\hat{\mathbf{j}} \times \mathbf{B} \Rightarrow \mathbf{B}=\hat{\mathbf{k}}$
Hence, magnetic field components,
$$ \mathbf{B}=2 \times 10^{-8} \hat{\mathbf{k}} T=2 \times 10^{-8} T $$
(along $z$-direction)
BETA
