A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the center of the coil is

Options

A. `2.01 xx 10^(-3)T`
B. `5.64 xx 10^(-3)T`
C. `2.64 xx 10^(-3)T`
D. `5.64 xx 10^(-3)T`

Explanation

To find the magnetic field at the center of a 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, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( n \) is the number of turns per unit length (in this case, total turns since the coil is circular),
- \( I \) is the current in amperes,
- \( R \) is the radius of the coil in meters.

**Step 1: Identify the given values.**
- Radius \( R = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m} \)
- Number of turns \( N = 100 \)
- Current \( I = 3.2 \, \text{A} \)

**Step 2: Calculate the number of turns per unit length.**
Since the coil is circular, the number of turns per unit length \( n \) is simply the total number of turns \( N \):
\[ n = N = 100 \]

**Step 3: Substitute the values into the formula.**
Now we can substitute the values into the formula for the magnetic field:
\[ B = \frac{(4\pi \times 10^{-7}) \times 100 \times 3.2}{2 \times 0.1} \]

**Step 4: Simplify the expression.**
Calculating the denominator:
\[ 2 \times 0.1 = 0.2 \]
Now substituting this back into the equation:
\[ B = \frac{(4\pi \times 10^{-7}) \times 100 \times 3.2}{0.2} \]

**Step 5: Calculate the numerator.**
Calculating the numerator:
\[ (4\pi \times 10^{-7}) \times 100 \times 3.2 = 1280\pi \times 10^{-7} \]

**Step 6: Calculate the magnetic field \( B \).**
Now, substituting the numerator into the equation:
\[ B = \frac{1280\pi \times 10^{-7}}{0.2} = 6400\pi \times 10^{-7} \]
Using \( \pi \approx 3.14 \):
\[ B \approx 6400 \times 3.14 \times 10^{-7} \approx 2000.8 \times 10^{-7} \, \text{T} \]
\[ B \approx 2.008 \times 10^{-3} \, \text{T} \approx 2.01 \times 10^{-3} \, \text{T} \]

Thus, the magnetic field at the center of the coil is approximately:
\[ B \approx 2.01 \times 10^{-3} \, \text{T} \]

**Final Answer:**
The magnetic field at the center of the coil is \( 2.01 \times 10^{-3} \, \text{T} \).

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