A current of `0.1 A` circulates around a coil of 100 turns and having a radius equal to `5 cm`. The magnetic field set up at the centre of the coil is
(`mu_(0)=4pixx10^(-7)` weber/amper-metre)

Options

A. `5pixx10^(-5)T`
B. `8pixx10^(-5)T`
C. `4pixx10^(-5)T`
D. `2pixx10^(-5)T`

Answer

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

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

Where:
- \( B \) is the magnetic field at the center of the coil,
- \( \mu_0 \) is the permeability of free space, given as \( 4\pi \times 10^{-7} \, \text{Wb/A·m} \),
- \( n \) is the number of turns per unit length (in this case, total turns since we are considering the entire coil),
- \( I \) is the current in amperes,
- \( R \) is the radius of the coil in meters.

### Step 1: Convert the radius from centimeters to meters
Given:
- Radius \( R = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)

### Step 2: Identify the values
- Current \( I = 0.1 \, \text{A} \)
- Number of turns \( n = 100 \)
- Permeability \( \mu_0 = 4\pi \times 10^{-7} \, \text{Wb/A·m} \)

### Step 3: Substitute the values into the formula
Now we can substitute the values into the formula for the magnetic field:

\[ B = \frac{4\pi \times 10^{-7} \cdot 100 \cdot 0.1}{2 \cdot (5 \times 10^{-2})} \]

### Step 4: Simplify the equation
Calculating the denominator:

\[ 2 \cdot (5 \times 10^{-2}) = 1 \times 10^{-1} = 0.1 \]

Now substituting this back into the equation:

\[ B = \frac{4\pi \times 10^{-7} \cdot 100 \cdot 0.1}{0.1} \]

### Step 5: Cancel out the \( 0.1 \) in the numerator and denominator
This simplifies to:

\[ B = 4\pi \times 10^{-7} \cdot 100 \]

### Step 6: Calculate the result
Now, calculating the multiplication:

\[ B = 4\pi \times 10^{-5} \, \text{T} \]

### Final Answer
Thus, the magnetic field at the center of the coil is:

\[ B = 4\pi \times 10^{-5} \, \text{T} \]