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Step-by-Step Solution
Step 1: Understand the Problem
We have a plane electromagnetic wave whose magnetic field is given by
$ \overline{B} = B_0 \,\hat{i}\,\cos(kz - \omega t) + B_1 \,\hat{j}\,\cos(kz + \omega t). $
The values of $B_0$ and $B_1$ are:
$ B_0 = 3 \times 10^{-5}\,\text{T} $
$ B_1 = 2 \times 10^{-6}\,\text{T} $
We need to find the root mean square (rms) force experienced by a stationary charge $Q = 10^{-4}\,\text{C}$ placed at $z=0$.
Step 2: Relate Magnetic Field to Electric Field in an Electromagnetic Wave
In a plane electromagnetic wave traveling in vacuum,
The electric field magnitude is related to the magnetic field magnitude by $ E = c\,B $, where $ c = 3 \times 10^8\,\text{m/s} $ is the speed of light in vacuum.
The directions of $ \overline{E} $ and $ \overline{B} $ are perpendicular, and both are perpendicular to the direction of propagation.
Since we have two components $B_0$ and $B_1$, their corresponding electric field amplitudes are:
$ \overrightarrow{E_0} = B_0 \, c \,(\text{direction}) \quad \text{and} \quad \overrightarrow{E_1} = B_1 \, c \,(\text{direction}). $
From the problem’s given orientation, we can treat:
$ E_0 = B_0 \, c = 3 \times 10^{-5} \times 3 \times 10^8 \,\text{V/m}, $
$ E_1 = B_1 \, c = 2 \times 10^{-6} \times 3 \times 10^8 \,\text{V/m}. $
The negative or positive directions indicated do not affect the magnitudes for force calculation; we will ultimately combine these field amplitudes vectorially.
Step 3: Find the Net Maximum Electric Field Amplitude
Because these two electric fields are perpendicular components, the net maximum electric field $E_{\text{net}}$ is given by the vector sum:
$ E_{\text{net}} = \sqrt{E_0^2 + E_1^2}. $
Substitute the values:
$ E_0 = 3 \times 10^{-5} \times 3 \times 10^8 = 9 \times 10^3 \,\text{V/m}, $
$ E_1 = 2 \times 10^{-6} \times 3 \times 10^8 = 6 \times 10^2 \,\text{V/m}. $
So,
$ E_{\text{net}} = \sqrt{\left(9 \times 10^3\right)^2 + \left(6 \times 10^2\right)^2}
= \sqrt{(81 \times 10^6) + (3.6 \times 10^5)} \approx \sqrt{81.36 \times 10^6}. $
$ E_{\text{net}} \approx 9.02 \times 10^3\,\text{V/m}. $
Step 4: Calculate the Maximum Force on the Charge
The instantaneous force on a stationary charge $Q$ due to the electric field is:
$ \overrightarrow{F} = Q \,\overrightarrow{E}.
$
The maximum value of the force (when the electric field is at its maximum amplitude) becomes:
$ F_{\text{max}} = Q \, E_{\text{net}}.
$
Substitute $ Q = 10^{-4}\,\text{C} $ and $ E_{\text{net}} \approx 9.02 \times 10^3\,\text{V/m}$:
$ F_{\text{max}} = 10^{-4} \times 9.02 \times 10^3 = 0.902\,\text{N} \approx 0.9\,\text{N}.
$
Step 5: Find the RMS Value of the Force
The force in an electromagnetic wave varies sinusoidally with time. The rms (root mean square) value of a sinusoidal quantity of peak value $F_{\text{max}}$ is:
$ F_{\text{rms}} = \frac{F_{\text{max}}}{\sqrt{2}}.
$
Hence,
$ F_{\text{rms}} = \frac{0.9\,\text{N}}{\sqrt{2}} \approx \frac{0.9}{1.414} \approx 0.636\,\text{N}.
$
Rounded suitably, this is about $0.6\,\text{N}$. Thus the closest answer is 0.6 N.
