$\frac{qva}{8}$
Step 1: The magnetic moment of a charged particle moving in a circular path is given by: \[ \mu = I A \] where $I$ is the current and $A$ is the area of the circular path.
Step 2: The current $I$ is given by: \[ I = \frac{q}{T} \] where $T$ is the time period of the circular motion. The time period is: \[ T = \frac{2\pi a}{v} \] Step 3: Substituting $T$: \[ I = \frac{q}{2\pi a / v} = \frac{q v}{2\pi a} \]
Step 4: The area of the circular path is: \[ A = \pi a^2 \]
Step 5: Compute the magnetic moment: \[ \mu = \left(\frac{q v}{2\pi a}\right) (\pi a^2) \] \[ = \frac{q v a}{2} \]
Step 6: Therefore, the correct answer is (C).
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: