A closed circuit of a resistor `R`, inductor of inductance `L` and a source of emf `E` are connected is series. If the inductance of the coil is abruptly decreaed to `L//4` (by removing its magnetic core), the new current immediately after this moment is (before decreasing the inductance the circuit is in steady state)

Options

A. zero
B. `(E)/(R )`
C. `4(E)/(R )`
D. `(E)/(4R)`

Explanation

To solve the problem, we need to analyze the circuit before and after the inductance change. Here’s a step-by-step solution:

### Step 1: Determine the initial current in the circuit
In a steady state, the current through the circuit can be calculated using Ohm's law. The total voltage in the circuit is equal to the EMF (E) provided by the source.

The initial current \( I_0 \) in the circuit can be given by:
\[ I_0 = \frac{E}{R} \]

### Step 2: Understand the effect of changing inductance
When the inductance \( L \) is abruptly decreased to \( \frac{L}{4} \), the inductor will try to maintain the current flowing through it due to its property of opposing changes in current.

### Step 3: Apply the inductor's property
The inductor's property states that the voltage across the inductor is given by:
\[ V_L = L \frac{di}{dt} \]
Since the inductance is changed abruptly, we can assume that the change in current \( di \) happens over an infinitesimally small time \( dt \).

### Step 4: Calculate the new current immediately after the change
At the moment the inductance changes, the inductor will still have the same current flowing through it as it did just before the change. Therefore, we can set up the equation:
\[ E = L \frac{di}{dt} + R I \]
However, since the inductance has changed to \( \frac{L}{4} \), we can express the voltage across the inductor just after the change as:
\[ E = \frac{L}{4} \frac{di}{dt} + R I \]
At the moment of change, \( I \) is still \( I_0 \), so we can substitute \( I_0 \) into the equation:
\[ E = \frac{L}{4} \cdot 0 + R I_0 \]
This means that the current immediately after the change remains the same as it was before the change, which is:
\[ I = I_0 = \frac{E}{R} \]

### Step 5: Conclusion
Thus, the new current immediately after the inductance is decreased to \( \frac{L}{4} \) is:
\[ I = \frac{E}{R} \]

### Final Answer
The new current immediately after the change is \( \frac{E}{R} \).