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Step-by-Step Solution
Step 1: Understand the Scenario and Relevant Formula
We have an electromagnetic wave whose magnetic field at any position and time is given by
$B = B_{0} \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz - \omega t)$.
Two charges, $q_{1} = 4\pi$ coulomb and $q_{2} = 2\pi$ coulomb, are located at
$(0,0,\frac{\pi}{k})$ and $(0,0,\frac{3\pi}{k})$, respectively, moving with velocity
$\vec{v} = 0.5\,c\,\hat{i}$ at $t = 0$.
The magnetic force on a charge $q$, moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$, is given by
$\vec{F} = q \,(\vec{v} \times \vec{B})$.
Step 2: Determine the Magnetic Field at Each Charge's Location
At time $t=0$, for the charge $q_1$ at $z = \frac{\pi}{k}$:
Inside the cosine term for $q_1$, we have
$\cos\!\bigl(k \cdot \frac{\pi}{k} - \omega \cdot 0\bigr) = \cos(\pi) = -1.$
Similarly, for charge $q_2$ at $z = \frac{3\pi}{k}$:
$\cos\!\bigl(k \cdot \frac{3\pi}{k} - \omega \cdot 0\bigr) = \cos(3\pi) = -1.$
Hence, at $t=0$, both charges experience the same value of
$\cos(\dots) = -1$, so
$B = B_{0} \,\frac{\hat{i} + \hat{j}}{\sqrt{2}} \times (-1) \;=\; -\,B_{0}\,\frac{\hat{i} + \hat{j}}{\sqrt{2}}.$
Step 3: Calculate the Force on Each Charge
1. Force on $q_{1}$:
Using $\vec{F}_{1} = q_{1}\,[\vec{v} \times \vec{B}]$, we substitute
$q_{1} = 4\pi$,
$\vec{v} = 0.5\,c\,\hat{i}$,
and
$\vec{B} = -\,B_{0}\,\frac{\hat{i} + \hat{j}}{\sqrt{2}}.$
So,
\[
\vec{F}_{1}
= 4\pi \,\Bigl[\,0.5\,c\,\hat{i} \,\times\,\Bigl(-\,B_{0}\,\frac{\hat{i} + \hat{j}}{\sqrt{2}}\Bigr)\Bigr].
\]
2. Force on $q_{2}$:
Similarly, for $q_{2} = 2\pi$,
\[
\vec{F}_{2}
= 2\pi \,\Bigl[\,0.5\,c\,\hat{i} \,\times\,\Bigl(-\,B_{0}\,\frac{\hat{i} + \hat{j}}{\sqrt{2}}\Bigr)\Bigr].
\]
Step 4: Simplify the Cross Product
Recall that $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}.$ Hence,
\[
\hat{i} \times \bigl(\hat{i} + \hat{j}\bigr)
= 0 + \bigl(\hat{i} \times \hat{j}\bigr)
= \hat{k}.
\]
Thus, each force expression simplifies in a similar manner. Notice that the factor $-1$ from the cosine function merely contributes a negative sign. The crucial point is that both charges see the same $\cos(\dots)$ value.
Overall, if we factor out the common terms, the only difference in $\vec{F}_{1}$ and $\vec{F}_{2}$ arises from whether the charge is $4\pi$ or $2\pi$.
Step 5: Take the Ratio of the Two Forces
Since all other factors (including $0.5\,c$, $B_{0}$, and the vector cross product) appear identically in both forces, the ratio
$\frac{F_{1}}{F_{2}}$
depends purely on the ratio of the charges:
\[
\frac{F_{1}}{F_{2}}
= \frac{4\pi}{2\pi} = 2 : 1.
\]
Step 6: Final Answer
The ratio of the force on $q_{1}$ to the force on $q_{2}$ is
$\boxed{2:1}.$
