A 100 turn closely wound circular coil of radius `10cm` carries a current of `3*2A`.What is the field at the centre of the coil?

Options

A. `2xx10T`
B. `2xx10^(-3)`T
C. `2xx10^(-6)` T
D. `2xx10^(-9)`T

Answer

To find the magnetic field at the center of a closely wound circular coil, we can use the formula:

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

where:
- \( B \) is the magnetic field at the center of the coil,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( n \) is the number of turns per unit length (in this case, total turns since the coil is closely wound),
- \( I \) is the current in amperes,
- \( r \) is the radius of the coil in meters.

### Step-by-Step Solution:

1. **Identify the given values**:
- Number of turns, \( n = 100 \)
- Radius of the coil, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current, \( I = 3.2 \, \text{A} \)

2. **Substitute the values into the formula**:
\[ B = \frac{\mu_0 n I}{2r} \]
Substituting the known values:
\[ B = \frac{(4\pi \times 10^{-7} \, \text{T m/A}) \times 100 \times 3.2}{2 \times 0.1} \]

3. **Calculate the numerator**:
\[ \text{Numerator} = 4\pi \times 10^{-7} \times 100 \times 3.2 \]
\[ = 4\pi \times 320 \times 10^{-7} \]
\[ = 1280\pi \times 10^{-7} \]

4. **Calculate the denominator**:
\[ \text{Denominator} = 2 \times 0.1 = 0.2 \]

5. **Combine the results**:
\[ B = \frac{1280\pi \times 10^{-7}}{0.2} \]
\[ = 6400\pi \times 10^{-7} \]

6. **Calculate the numerical value**:
Using \( \pi \approx 3.14 \):
\[ B \approx 6400 \times 3.14 \times 10^{-7} \]
\[ \approx 20000.8 \times 10^{-7} \, \text{T} \]
\[ = 2 \times 10^{-3} \, \text{T} \]

### Final Answer:
The magnetic field at the center of the coil is approximately \( 2 \times 10^{-3} \, \text{T} \).

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