When the current in a certain inductor coil is `5.0 A` and is increasing at the rate of `10.0 A//s`,the magnitude of potential difference across the coil is `140 V`.When the current is `5.0 A` and decreasing at the rate of `10.0 A//s`,the potential difference is `60 V`.The self inductance of the coil is :

Options

A. `2H`
B. `4H`
C. `10H`
D. `12H`

Answer

To solve the problem, we will analyze the two cases provided and apply the formula for the potential difference across an inductor.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have an inductor coil with a current of 5.0 A. The current is increasing at a rate of 10.0 A/s in the first case, resulting in a potential difference of 140 V. In the second case, the current is decreasing at the same rate of 10.0 A/s, leading to a potential difference of 60 V. We need to find the self-inductance \( L \) of the coil.

2. **Using the Formula for Inductance**:
The potential difference \( e \) across an inductor can be expressed as:
\[ e = R \cdot I + L \cdot \frac{dI}{dt} \]
where \( R \) is the resistance, \( I \) is the current, and \( \frac{dI}{dt} \) is the rate of change of current.

3. **Case 1 (Current Increasing)**:
- Given:
- \( I = 5.0 \, \text{A} \)
- \( \frac{dI}{dt} = 10.0 \, \text{A/s} \)
- \( e = 140 \, \text{V} \)
- Substituting into the formula:
\[ 140 = R \cdot 5 + L \cdot 10 \quad \text{(Equation 1)} \]

4. **Case 2 (Current Decreasing)**:
- Given:
- \( I = 5.0 \, \text{A} \)
- \( \frac{dI}{dt} = -10.0 \, \text{A/s} \)
- \( e = 60 \, \text{V} \)
- Substituting into the formula:
\[ 60 = R \cdot 5 - L \cdot 10 \quad \text{(Equation 2)} \]

5. **Solving the Equations**:
- From Equation 1:
\[ 140 = 5R + 10L \quad \text{(1)} \]
- From Equation 2:
\[ 60 = 5R - 10L \quad \text{(2)} \]
- Now, we can add both equations to eliminate \( L \):
\[ (140 + 60) = (5R + 10L) + (5R - 10L) \]
\[ 200 = 10R \]
\[ R = 20 \, \Omega \]

6. **Substituting \( R \) Back to Find \( L \)**:
- Substitute \( R = 20 \, \Omega \) into Equation 1:
\[ 140 = 5 \cdot 20 + 10L \]
\[ 140 = 100 + 10L \]
\[ 10L = 140 - 100 \]
\[ 10L = 40 \]
\[ L = 4 \, \text{H} \]

### Final Answer:
The self-inductance of the coil is \( L = 4 \, \text{H} \).