2A current is flowing in a circular coil of 'n' turn, having 10cm radius. The flux density at its centre is `0.126 xx 10^(-2) wb//m^(2)`. The value of 'n' is

Options

A. 10
B. 100
C. 1000
D. 1

Answer

To solve the problem of finding the number of turns 'n' in a circular coil given the current, radius, and magnetic flux density at its center, we can follow these steps:

### Step 1: Write down the formula for magnetic field (B) at the center of a circular coil.
The magnetic field at the center of a circular coil is given by the formula:

\[ B = \frac{n \mu_0 I}{2r} \]

Where:
- \( B \) = magnetic flux density (in Tesla)
- \( n \) = number of turns in the coil
- \( \mu_0 \) = permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \))
- \( I \) = current (in Amperes)
- \( r \) = radius of the coil (in meters)

### Step 2: Substitute the known values into the formula.
From the problem:
- \( B = 0.126 \times 10^{-2} \, \text{T} = 0.00126 \, \text{T} \)
- \( I = 2 \, \text{A} \)
- \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)

Now substituting these values into the formula:

\[ 0.00126 = \frac{n \cdot (4\pi \times 10^{-7}) \cdot 2}{2 \cdot 0.1} \]

### Step 3: Simplify the equation.
The equation simplifies to:

\[ 0.00126 = \frac{n \cdot (4\pi \times 10^{-7}) \cdot 2}{0.2} \]

This can be further simplified to:

\[ 0.00126 = n \cdot (4\pi \times 10^{-7} \cdot 10) \]

\[ 0.00126 = n \cdot (4\pi \times 10^{-6}) \]

### Step 4: Solve for 'n'.
Now, we can isolate 'n':

\[ n = \frac{0.00126}{4\pi \times 10^{-6}} \]

Calculating the denominator:

\[ 4\pi \approx 12.56 \quad \text{(using } \pi \approx 3.14\text{)} \]

Thus,

\[ n = \frac{0.00126}{12.56 \times 10^{-6}} \approx \frac{0.00126}{0.00001256} \approx 100.32 \]

### Step 5: Round to the nearest whole number.
Since 'n' must be a whole number, we round 100.32 to 100.

### Conclusion:
The value of 'n' is approximately 100 turns.

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