Electrostatics 7 Question 1
1. The magnetic field of a plane electromagnetic wave is given by
$\mathbf{B}=B_{0}[\cos (k z-\omega t)] \hat{\mathbf{i}}+B_{1} \cos (k z+\omega t) \hat{\mathbf{j}}$
where, $B_{0}=3 \times 10^{-5} \mathrm{~T}$ and $B_{1}=2 \times 10^{-6} \mathrm{~T}$. The rms value of the force experienced by a stationary charge $Q=10^{-4} \mathrm{C}$ at $z=0$ is closest to
(Main 2019, 9 April I)
(a) $0.1 \mathrm{~N}$
(b) $3 \times 10^{-2} \mathrm{~N}$
(c) $0.6 \mathrm{~N}$
(d) $0.9 \mathrm{~N}$
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Answer:
Correct Answer: 1. (c)
Solution:
- Given, magnetic field of an electromagnetic wave is
$$ \mathbf{B}=B_{0}\left[\cos (k z-\omega t] \hat{\mathbf{i}}+B_{1}[\cos (k z+\omega t] \hat{\mathbf{j}}\right. $$
Here, $B_{0}=3 \times 10^{-5} \mathrm{~T}$ and $B_{1}=2 \times 10^{-6} \mathrm{~T}$
Also, stationary charge, $Q=10^{-4} \mathrm{C}$ at $z=0$
As charge is released from the rest at $z=0$, in this condition.
Maximum electric field, $E_{0}=c B_{0}$ and $E_{1}=c B_{1}$
So, $E_{0}=c \times 3 \times 10^{-5}$ and $E_{1}=c \times 2 \times 10^{-6}$
Now,the direction of electric field of an electromagnetic wave is perpendicular to $\mathbf{B}$ and to the direction of propagation of wave $(\mathbf{E} \times \mathbf{B})$ which is $\hat{\mathbf{k}}$
So, for $E_{0}, \mathbf{E} _ {0} \times \mathbf{B} _ {0}=\hat{\mathbf{k}} \Rightarrow \mathbf{E} _ {0} \times \hat{\mathbf{i}}=\hat{\mathbf{k}} \Rightarrow \mathbf{E} _ {0}=-\hat{\mathbf{j}}$
Similarly,
$$ \begin{aligned} & \text { for } E_{1} \mathbf{E} _ {1} \times \mathbf{B} _ {1}=\mathbf{k} \Rightarrow \mathbf{E} _ {1} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} \\ & \Rightarrow \quad \mathbf{E} _ {1}=\hat{\mathbf{i}} \\ & \therefore \quad \mathbf{E} _ {0}=c \times 3 \times 10^{-5}(-\hat{\mathbf{j}}) \mathrm{NC}^{-1} \\ & \mathbf{E}=c \times 2 \times 10^{-6}(+\hat{\mathbf{i}}) \mathrm{NC}^{-1} \end{aligned} $$
$\therefore$ Maximum force experienced by stationary charge is
$$ \begin{aligned} \mathbf{F} _ {\max } & =Q \mathbf{E}=Q\left(\mathbf{E} _ {0}+\mathbf{E} _ {1}\right) \\ & =Q \times c\left[-3 \times 10^{-5} \hat{\mathbf{j}}+2 \times 10^{-6} \hat{\mathbf{i}}\right] \end{aligned} $$
$$ \begin{aligned} \Rightarrow\left|\mathbf{F} _ {\max }\right| & =10^{-4} \times 3 \times 10^{8} \times \sqrt{\left(3 \times 10^{-5}\right)^{2}+\left(2 \times 10^{-6}\right)^{2}} \\ & =3 \times 10^{4} \times 10^{-6} \sqrt{900+4} \\ & =3 \times 10^{-2} \times \sqrt{904} \approx 0.9 \mathrm{~N} \end{aligned} $$
$\therefore$ rms value of experienced force is
$$ \begin{aligned} F_{\mathrm{rms}} & =\frac{F_{\max }}{\sqrt{2}}=\frac{0.9}{\sqrt{2}}=0.707 \times 0.9 \\ & =0.6363 \mathrm{~N} \approx 0.6 \mathrm{~N} \end{aligned} $$
