A superconducting loop of radius `R` has self inductance `L`.`A` uniform & constant magnetic field `B` is applied perpendicular to the plane of the loop. Initially current in this loop is zero. The loop is rotated about its diameter by `180^(@)`.Find the current in the loop after rotation.

Options

A. zero
B. `(BpiR^(2))/(L)`
C. `(2BpiR^(2))/(L)`
D. `(BpiR^(2))/(2L)`

Answer

To solve the problem, we need to find the current induced in a superconducting loop when it is rotated by 180 degrees in a uniform magnetic field. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
- We have a superconducting loop with radius \( R \) and self-inductance \( L \).
- A uniform magnetic field \( B \) is applied perpendicular to the plane of the loop.
- Initially, the current in the loop is zero.

### Step 2: Calculate the Initial Magnetic Flux
- The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the loop.
- The area \( A \) of the loop is:
\[ A = \pi R^2 \]
- Therefore, the initial magnetic flux \( \Phi_i \) is:
\[ \Phi_i = B \cdot \pi R^2 \]

### Step 3: Consider the Final Magnetic Flux After Rotation
- When the loop is rotated by 180 degrees, the angle between the magnetic field and the area vector changes from \( 0^\circ \) to \( 180^\circ \).
- The final magnetic flux \( \Phi_f \) is given by:
\[ \Phi_f = B \cdot A \cdot \cos(180^\circ) = -B \cdot \pi R^2 \]

### Step 4: Calculate the Change in Magnetic Flux
- The change in magnetic flux \( \Delta \Phi \) is:
\[ \Delta \Phi = \Phi_f - \Phi_i = -B \cdot \pi R^2 - B \cdot \pi R^2 = -2B \cdot \pi R^2 \]

### Step 5: Apply Faraday's Law of Electromagnetic Induction
- According to Faraday's law, the induced electromotive force (emf) \( \mathcal{E} \) in the loop is related to the rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
- Since the loop is rotated instantaneously, we can consider the induced emf to be:
\[ \mathcal{E} = -\Delta \Phi = 2B \cdot \pi R^2 \]

### Step 6: Relate Induced emf to Current
- For a superconducting loop, the resistance \( R \) is zero, so the induced emf is related to the current \( I \) and the self-inductance \( L \) by:
\[ \mathcal{E} = L \frac{dI}{dt} \]
- Since the change is instantaneous, we can relate the induced emf directly to the current:
\[ \mathcal{E} = L I \]

### Step 7: Solve for the Induced Current
- Setting the two expressions for emf equal gives:
\[ L I = 2B \cdot \pi R^2 \]
- Solving for the current \( I \):
\[ I = \frac{2B \cdot \pi R^2}{L} \]

### Final Answer
The current in the loop after rotation is:
\[ I = \frac{2B \cdot \pi R^2}{L} \]