The magnitude of the magnetic field at the centre of the tightly wound 150 turn coil of radius 12 cm carrying a current of 2 A is

Options

A. 18G
B. 19.7G
C. 15.7G
D. 17.7G

Explanation

To find the magnitude of the magnetic field at the center of a tightly wound coil, we can use the formula for the magnetic field \( B \) at the center of a circular coil:

\[ B = \frac{\mu_0 N I}{2R} \]

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( N \) is the number of turns in the coil,
- \( I \) 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 \( N = 150 \)
- Radius \( R = 12 \, \text{cm} = 12 \times 10^{-2} \, \text{m} \)
- Current \( I = 2 \, \text{A} \)

2. **Substitute the values into the formula**:
\[ B = \frac{\mu_0 N I}{2R} \]
Substituting the known values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 150 \times 2}{2 \times (12 \times 10^{-2})} \]

3. **Simplify the expression**:
- First, calculate \( 2R \):
\[ 2R = 2 \times (12 \times 10^{-2}) = 24 \times 10^{-2} = 0.24 \, \text{m} \]
- Now substitute this back into the equation:
\[ B = \frac{(4\pi \times 10^{-7}) \times 150 \times 2}{0.24} \]

4. **Calculate the numerator**:
- Calculate \( 4\pi \times 150 \times 2 \):
\[ 4\pi \times 150 \times 2 = 1200\pi \approx 3769.91 \]
- Therefore:
\[ B = \frac{3769.91 \times 10^{-7}}{0.24} \]

5. **Calculate the final value of B**:
- Divide \( 3769.91 \times 10^{-7} \) by \( 0.24 \):
\[ B \approx 15624.63 \times 10^{-7} \, \text{T} = 1.562463 \times 10^{-3} \, \text{T} \]

6. **Convert to Gauss**:
- Since \( 1 \, \text{T} = 10^4 \, \text{G} \):
\[ B \approx 1.562463 \times 10^{-3} \times 10^4 \, \text{G} = 15.62463 \, \text{G} \]
- Rounding to two decimal places gives:
\[ B \approx 15.7 \, \text{G} \]

### Final Answer:
The magnitude of the magnetic field at the center of the coil is approximately **15.7 G**.