If in a coil rate of change of area is `(5meter^(2))/(milli second)`and current become `1 amp` from `2 amp` in `2xx10^(-3)` sec. if magnetic field is `1` Tesla then self-inductance of the coil is

Options

A. `2H`
B. `5H`
C. `20 H`
D. `10 H`

Answer

To solve the problem step by step, we will use the information provided and apply the relevant equations from electromagnetic theory.

### Step 1: Identify the given values
- Rate of change of area, \( \frac{dA}{dt} = 5 \, \text{m}^2/\text{ms} = 5 \times 10^3 \, \text{m}^2/\text{s} \)
- Change in current, \( \Delta I = 2 \, \text{A} - 1 \, \text{A} = 1 \, \text{A} \)
- Time interval for the change in current, \( \Delta t = 2 \times 10^{-3} \, \text{s} \)
- Magnetic field, \( B = 1 \, \text{T} \)

### Step 2: Calculate the rate of change of current
The rate of change of current \( \frac{dI}{dt} \) can be calculated as:
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{1 \, \text{A}}{2 \times 10^{-3} \, \text{s}} = 500 \, \text{A/s} \]

### Step 3: Use the relationship between magnetic flux, self-inductance, and current
The relationship we will use is:
\[ \frac{d\Phi}{dt} = L \frac{dI}{dt} \]
where \( \Phi \) is the magnetic flux.

### Step 4: Express magnetic flux in terms of area and magnetic field
The magnetic flux \( \Phi \) can be expressed as:
\[ \Phi = B \cdot A \]
Thus, the rate of change of magnetic flux is:
\[ \frac{d\Phi}{dt} = B \frac{dA}{dt} \]

### Step 5: Substitute values into the equation
Now we can substitute the values into the equation:
\[ B \frac{dA}{dt} = L \frac{dI}{dt} \]
Substituting \( B = 1 \, \text{T} \), \( \frac{dA}{dt} = 5 \times 10^3 \, \text{m}^2/\text{s} \), and \( \frac{dI}{dt} = 500 \, \text{A/s} \):
\[ 1 \cdot (5 \times 10^3) = L \cdot (500) \]

### Step 6: Solve for self-inductance \( L \)
Rearranging the equation gives:
\[ L = \frac{5 \times 10^3}{500} = 10 \, \text{H} \]

### Final Answer
The self-inductance of the coil is \( L = 10 \, \text{H} \).

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