Consider the situation shown in the figure of the previous problem. Suppose the wire connecting O and C has zero resistance but the circular loop has a resistance R uniformly distributed along its length. The rod OA is made to rotate with a uniform angular speed `(omega)` as shown in the figure. Find the current in the rod when `-AOC = 90^(@)`

Options

A. `5/3 (Ba^(2)omega)/(R)`
B. `8/3 (Ba^(2)omega)/(R)`
C. `(Ba^(2)omega)/(3R)`
D. None of these

Explanation

To solve the problem, we need to find the current in the rod when the angle AOC is 90 degrees. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a circular loop with a resistance \( R \) uniformly distributed along its length. The rod \( OA \) is rotating with a uniform angular speed \( \omega \). When \( \angle AOC = 90^\circ \), the rod is perpendicular to the radius at point A.

### Step 2: Determine the Length of the Rod
When \( \angle AOC = 90^\circ \), the length of the rod \( OA \) can be denoted as \( L \). The length of the circular arc \( AB \) can be expressed as:
\[ L_{AB} = \frac{1}{4} \times 2\pi r = \frac{\pi r}{2} \]
where \( r \) is the radius of the circular loop.

### Step 3: Calculate the Resistance of the Arc
The total resistance \( R \) is uniformly distributed along the circular loop. The resistance of the arc \( AB \) can be calculated as:
\[ R_{AB} = \frac{R}{\text{Total Length}} \times \text{Length of Arc AB} = \frac{R}{2\pi r} \times \frac{\pi r}{2} = \frac{R}{4} \]
The remaining arc \( AOB \) has a length of \( \frac{3\pi r}{2} \), so its resistance is:
\[ R_{AOB} = \frac{R}{2\pi r} \times \frac{3\pi r}{2} = \frac{3R}{4} \]

### Step 4: Find the Equivalent Resistance
Since the resistances \( R_{AB} \) and \( R_{AOB} \) are in parallel, we can find the equivalent resistance \( R_{eq} \) using the formula for resistors in parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{R_{AB}} + \frac{1}{R_{AOB}} = \frac{1}{\frac{R}{4}} + \frac{1}{\frac{3R}{4}} = \frac{4}{R} + \frac{4}{3R} = \frac{12 + 4}{3R} = \frac{16}{3R} \]
Thus, the equivalent resistance is:
\[ R_{eq} = \frac{3R}{16} \]

### Step 5: Calculate the Induced EMF
The induced EMF \( \mathcal{E} \) in the rod can be calculated using the formula:
\[ \mathcal{E} = B \cdot L \cdot v \]
where \( v \) is the linear velocity of the end of the rod, given by \( v = \frac{a}{2} \omega \). Therefore, the EMF becomes:
\[ \mathcal{E} = B \cdot L \cdot \left(\frac{a}{2} \omega\right) = B \cdot a \cdot \left(\frac{a}{2} \omega\right) = \frac{1}{2} B a^2 \omega \]

### Step 6: Calculate the Current
Using Ohm's law, the current \( I \) in the circuit can be calculated as:
\[ I = \frac{\mathcal{E}}{R_{eq}} = \frac{\frac{1}{2} B a^2 \omega}{\frac{3R}{16}} = \frac{8}{3} \frac{B a^2 \omega}{R} \]

### Final Answer
Thus, the current in the rod when \( \angle AOC = 90^\circ \) is:
\[ I = \frac{8}{3} \frac{B a^2 \omega}{R} \]