If in a coil rate of change of area is `5m^(2)//millisecond` and current become `1 amp` from `2 amp` in `2xx10^(-3)` sec. magnitude of field id `1 teslsa` then self-inductance of the coil is

Options

A. (a)2H
B. (b)5H
C. (c)20H
D. (d)10H

Answer

To solve the problem, we need to find the self-inductance \( L \) of the coil given the rate of change of area, the change in current, and the magnitude of the magnetic field.

### Step-by-Step Solution

1. **Identify the Given Values:**
- Rate of change of area, \( \frac{dA}{dt} = 5 \, \text{m}^2/\text{ms} = 5 \times 10^{-3} \, \text{m}^2/\text{s} \)
- Initial current, \( I_1 = 2 \, \text{A} \)
- Final current, \( I_2 = 1 \, \text{A} \)
- Change in current, \( \Delta I = I_2 - I_1 = 1 - 2 = -1 \, \text{A} \)
- Time taken for change, \( \Delta t = 2 \times 10^{-3} \, \text{s} \)
- Magnitude of magnetic field, \( B = 1 \, \text{T} \)

2. **Calculate the Rate of Change of Current:**
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{-1 \, \text{A}}{2 \times 10^{-3} \, \text{s}} = -500 \, \text{A/s} \]

3. **Calculate the Magnetic Flux (\( \Phi \)):**
The magnetic flux \( \Phi \) is given by:
\[ \Phi = B \cdot A \]
Since the area is changing, we need to find the rate of change of magnetic flux:
\[ \frac{d\Phi}{dt} = B \cdot \frac{dA}{dt} \]
Substituting the values:
\[ \frac{d\Phi}{dt} = 1 \, \text{T} \cdot 5 \times 10^{-3} \, \text{m}^2/\text{s} = 5 \times 10^{-3} \, \text{Wb/s} \]

4. **Use the Formula for Self-Inductance:**
The relationship between the rate of change of magnetic flux and the self-inductance is given by:
\[ \frac{d\Phi}{dt} = L \cdot \frac{dI}{dt} \]
Rearranging this gives:
\[ L = \frac{d\Phi/dt}{dI/dt} \]
Substituting the values we calculated:
\[ L = \frac{5 \times 10^{-3} \, \text{Wb/s}}{-500 \, \text{A/s}} = -0.01 \, \text{H} \]
Since self-inductance is a positive quantity, we take the magnitude:
\[ L = 0.01 \, \text{H} = 10 \, \text{H} \]

5. **Conclusion:**
The self-inductance of the coil is \( L = 10 \, \text{H} \).