A magnetic flux of `9 xx 10^(-4)` weber is linked with each turn of a 200 turn coil when there is an electric current of 3 A in it. The self inductance of the coil is

Options

A. `3 xx 10^(-2)` H
B. ` 6xx 10^(-2)` H
C. `4 xx 10^(-2)` H
D. `10^(-3)` H

Answer

To find the self-inductance of the coil, we can use the formula that relates magnetic flux (Φ), self-inductance (L), current (I), and the number of turns (N) in the coil. The formula is given by:

\[ \Phi = L \cdot I \]

Where:
- \(\Phi\) is the magnetic flux linked with each turn of the coil,
- \(L\) is the self-inductance of the coil,
- \(I\) is the current flowing through the coil.

Since we have a coil with multiple turns, the total magnetic flux linked with the coil can be expressed as:

\[ \Phi_{\text{total}} = N \cdot \Phi \]

Where:
- \(N\) is the number of turns in the coil.

Given:
- \(\Phi = 9 \times 10^{-4} \, \text{weber}\)
- \(N = 200\)
- \(I = 3 \, \text{A}\)

### Step 1: Calculate the total magnetic flux linked with the coil.

\[ \Phi_{\text{total}} = N \cdot \Phi = 200 \cdot (9 \times 10^{-4}) = 1.8 \times 10^{-1} \, \text{weber} \]

### Step 2: Use the formula to find self-inductance (L).

Rearranging the formula \(\Phi = L \cdot I\) gives us:

\[ L = \frac{\Phi_{\text{total}}}{I} \]

Substituting the values we have:

\[ L = \frac{1.8 \times 10^{-1}}{3} \]

### Step 3: Perform the calculation.

\[ L = 0.06 \, \text{H} = 6 \times 10^{-2} \, \text{H} \]

### Final Answer:

The self-inductance of the coil is \(6 \times 10^{-2} \, \text{Henry}\).

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