Steady current i flowing through a solenoid, whose core is made with a material of permeability `mu`. If core of the solenoid is abruptly removed then what current will flow through the circuit just after core is removed.

Options

A. `mui_(0)`
B. `(i_(0))/(mu)`
C. `i_(0)`
D. 0

Explanation

To solve the problem of determining the current flowing through a circuit just after the core of a solenoid is removed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
- A steady current \( i_0 \) is flowing through the solenoid when the core (with permeability \( \mu \)) is present. The self-inductance of the solenoid is \( L \).

2. **Identify the Change When the Core is Removed**:
- Upon removing the core, the self-inductance changes. The new self-inductance \( L' \) of the solenoid without the core can be expressed as:
\[ L' = \frac{L}{\mu} \]

3. **Magnetic Flux Relation**:
- The magnetic flux \( \Phi \) through the solenoid is related to the self-inductance and the current:
\[ \Phi = L \cdot i \]
- Initially, with the core, the magnetic flux is:
\[ \Phi_1 = L \cdot i_0 \]
- After removing the core, the magnetic flux is:
\[ \Phi_2 = L' \cdot i \]

4. **Equate the Magnetic Flux Before and After**:
- Since the magnetic flux must remain continuous (i.e., \( \Phi_1 = \Phi_2 \)), we can set up the equation:
\[ L \cdot i_0 = L' \cdot i \]

5. **Substitute the Expression for \( L' \)**:
- Substitute \( L' \) into the equation:
\[ L \cdot i_0 = \left(\frac{L}{\mu}\right) \cdot i \]

6. **Solve for the Current \( i \)**:
- Rearranging the equation gives:
\[ i = \frac{L \cdot i_0 \cdot \mu}{L} \]
- Simplifying this, we find:
\[ i = \mu \cdot i_0 \]

### Final Result:
The current flowing through the circuit just after the core is removed is:
\[ i = \mu \cdot i_0 \]