A circular loop of metal wire of radius r is placed perpendicular to uniform magnetic field B. Half of the loop is folded about the diameter with constant angular velocity `omega`. If resistance of the loop is R then current in the loop is

Options

A. zero, when `theta = 0`
B. `(pir^(2)Bomega)/(2R)` when `theta = 0`
C. `(pir^(2)Bomega)/(2R)` when `theta = (pi)/(6)`
D. `(pir^(2)Bomega)/(4R)` when `theta = (pi)/(6)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understanding the Setup
We have a circular loop of radius \( r \) placed perpendicular to a uniform magnetic field \( B \). The loop is folded about its diameter with a constant angular velocity \( \omega \).

### Step 2: Determine the Area of the Loop
Initially, the area \( A \) of the circular loop is given by:
\[ A = \pi r^2 \]
Since half of the loop is folded, the effective area that contributes to the magnetic flux will change as the loop rotates.

### Step 3: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the loop at any angle \( \theta \) can be expressed as:
\[ \Phi = B \cdot A_{\text{effective}} = B \cdot \left(\frac{\pi r^2}{2}\right) \cdot \cos(\theta) \]
where \( A_{\text{effective}} \) is the area of the half loop that is still exposed to the magnetic field.

### Step 4: Determine the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) in the loop is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Substituting the expression for magnetic flux:
\[ \mathcal{E} = -\frac{d}{dt}\left(B \cdot \frac{\pi r^2}{2} \cdot \cos(\theta)\right) \]
Using the chain rule, we have:
\[ \mathcal{E} = -B \cdot \frac{\pi r^2}{2} \cdot \frac{d}{dt}(\cos(\theta)) = -B \cdot \frac{\pi r^2}{2} \cdot (-\sin(\theta) \cdot \frac{d\theta}{dt}) \]
Since \( \frac{d\theta}{dt} = \omega \):
\[ \mathcal{E} = \frac{\pi r^2 B \omega}{2} \sin(\theta) \]

### Step 5: Calculate the Current in the Loop
Using Ohm's law, the current \( I \) in the loop can be calculated as:
\[ I = \frac{\mathcal{E}}{R} = \frac{\frac{\pi r^2 B \omega}{2} \sin(\theta)}{R} \]
Thus, the current in the loop is:
\[ I = \frac{\pi r^2 B \omega}{2R} \sin(\theta) \]

### Final Result
The current in the loop as it is folded and rotated is given by:
\[ I = \frac{\pi r^2 B \omega}{2R} \sin(\theta) \]

---