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, $B _0=3 \times 10^{-5} T$ and $B _1=2 \times 10^{-6} T$. The rms value of the force experienced by a stationary charge $Q=10^{-4} C$ at $z=0$ is closest to
(Main 2019, 9 April I)
(a) $0.1 N$
(b) $3 \times 10^{-2} N$
(c) $0.6 N$
(d) $0.9 N$
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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, $B _0=3 \times 10^{-5} T$ and $B _1=2 \times 10^{-6} T$
Also, stationary charge, $Q=10^{-4} C$ 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,
$$ \begin{aligned} & \text { for } E _1 \mathbf{E} _1 \times \mathbf{B} _1=\mathbf{k} \Rightarrow \mathbf{E} _1 \times \hat{\mathbf{j}}=\hat{\mathbf{k}} \\ & \Rightarrow \quad \mathbf{E} _1=\hat{\mathbf{i}} \\ & \therefore \quad \mathbf{E} _0=c \times 3 \times 10^{-5}(-\hat{\mathbf{j}}) NC^{-1} \\ & \mathbf{E}=c \times 2 \times 10^{-6}(+\hat{\mathbf{i}}) NC^{-1} \end{aligned} $$
$\therefore$ Maximum force experienced by stationary charge is
$$ \begin{aligned} \mathbf{F} _{\max } & =Q \mathbf{E}=Q\left(\mathbf{E} _0+\mathbf{E} _1\right) \\ & =Q \times c\left[-3 \times 10^{-5} \hat{\mathbf{j}}+2 \times 10^{-6} \hat{\mathbf{i}}\right] \end{aligned} $$
$$ \begin{aligned} \Rightarrow\left|\mathbf{F} _{\max }\right| & =10^{-4} \times 3 \times 10^{8} \times \sqrt{\left(3 \times 10^{-5}\right)^{2}+\left(2 \times 10^{-6}\right)^{2}} \\ & =3 \times 10^{4} \times 10^{-6} \sqrt{900+4} \\ & =3 \times 10^{-2} \times \sqrt{904} \approx 0.9 N \end{aligned} $$
$\therefore$ rms value of experienced force is
$$ \begin{aligned} F _{rms} & =\frac{F _{\max }}{\sqrt{2}}=\frac{0.9}{\sqrt{2}}=0.707 \times 0.9 \\ & =0.6363 N \approx 0.6 N \end{aligned} $$
