Modern Physics Ques 258
- The magnetic field of an electromagnetic wave is given by $\mathbf{B}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) Wbm^{-2}$
The associated electric field will be
(Main 2019, 8 April II)
(a) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(-2 \hat{\mathbf{j}}+\hat{\mathbf{i}}) Vm^{-1}$
(b) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(2 \hat{\mathbf{j}}+\hat{\mathbf{i}}) Vm^{-1}$
(c) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}) Vm^{-1}$
(d) $\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) Vm^{-1}$
Show Answer
Answer:
Correct Answer: 258.(c)
Solution:
Formula:
Relation Between The Magnetic Field Vector And The Electric Field Vector:
- Given,
$\mathbf{B}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})$ $ Wbm^{-2}$
From the given equation, it can be said that the electromagnetic wave is propagating negative $z$-direction, i.e. $-\hat{\mathbf{k}}$
Equation of associated electric field will be
$ \mathbf{E}=(|\mathbf{B}| c) \cos (k z+\omega t) \cdot \hat{\mathbf{n}} $
where, $\hat{\mathbf{n}}=$ a vector perpendicular to $\mathbf{B}$.
So, $|\mathbf{E}|=|\mathbf{B}| \cdot c$
$ =1.6 \times 10^{-6} \times 3 \times 10^{8}=4.8 \times 10^{2} $ $V / m $
Since, we know that for an electromagnetic wave, $\mathbf{E}$ and $\mathbf{B}$ are mutually perpendicular to each other.
So, $\mathbf{E} \cdot \mathbf{B}=0$
From the given options, when $\hat{\mathbf{n}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}$
$ \mathbf{E} \cdot \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})=0 $
Also, when $\hat{\mathbf{n}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$
$ \mathbf{E} \cdot \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}})=0 $
But, we also know that the direction of propagation of electromagnetic wave is perpendicular to both $\mathbf{E}$ and $\mathbf{B}$, i.e. it is in the direction of $\mathbf{E} \times \mathbf{B}$.
Again, when $\hat{\mathbf{n}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}$
$ \mathbf{E} \times \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})=-\hat{\mathbf{k}} $
and when $\hat{\mathbf{n}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$
$ \mathbf{E} \times \mathbf{B}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}})=\hat{\mathbf{k}} $
But, it is been given in the question that the direction of propagation of wave is in $-\hat{\mathbf{k}}$.
Thus, associated electric field will be
$\mathbf{E}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}) Vm^{-1}$
BETA
