the inductance of a closed-packed coil of `400` turns is `8mH`. `A` current of `5mA` is passed through it. The magnetic flux through each turn of the coil is

Options

A. `0.1muWb`
B. 0.2`mu`, Wb
C. `1.0mu` Wb
D. 2.0`mu` Wb

Explanation

To find the magnetic flux through each turn of the coil, we can follow these steps:

### Step 1: Understand the relationship between inductance, current, and magnetic flux.
The magnetic flux linkage (Ψ) in a coil is given by the product of its inductance (L) and the current (I) flowing through it:
\[ \Psi = L \cdot I \]

### Step 2: Write the formula for magnetic flux through each turn.
The total magnetic flux linkage can also be expressed in terms of the magnetic flux (φ) through each turn and the number of turns (n):
\[ \Psi = n \cdot φ \]
From this, we can derive the magnetic flux through each turn:
\[ φ = \frac{\Psi}{n} \]

### Step 3: Substitute the values into the equations.
Given:
- Inductance, \( L = 8 \, \text{mH} = 8 \times 10^{-3} \, \text{H} \)
- Current, \( I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A} \)
- Number of turns, \( n = 400 \)

First, calculate the magnetic flux linkage:
\[ \Psi = L \cdot I = (8 \times 10^{-3} \, \text{H}) \cdot (5 \times 10^{-3} \, \text{A}) = 40 \times 10^{-6} \, \text{Wb} = 4 \times 10^{-5} \, \text{Wb} \]

### Step 4: Calculate the magnetic flux through each turn.
Now, substitute the values into the equation for magnetic flux:
\[ φ = \frac{\Psi}{n} = \frac{4 \times 10^{-5} \, \text{Wb}}{400} = 1 \times 10^{-7} \, \text{Wb} \]

### Step 5: Convert the result into micro Weber.
Since \( 1 \, \text{Wb} = 10^{6} \, \mu \text{Wb} \):
\[ φ = 1 \times 10^{-7} \, \text{Wb} = 0.1 \, \mu \text{Wb} \]

### Final Answer:
The magnetic flux through each turn of the coil is \( 0.1 \, \mu \text{Wb} \).

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