What will be magnetic field at centre of current carrying circular loop of radius R?

Options

A. `(mu_(0)I)/(4piR)`
B. `(mu_(0)I)/(2piR)`
C. `(mu_(0)I)/(2R)`
D. zero

Answer

To find the magnetic field at the center of a current-carrying circular loop of radius \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circular loop of radius \( R \) carrying a current \( I \). We need to determine the magnetic field \( B \) at the center of this loop.

2. **Using Biot-Savart Law**:
- The magnetic field \( dB \) due to a small segment of the loop carrying current \( I \) can be expressed using the Biot-Savart Law:
\[ dB = \frac{\mu_0}{4\pi} \frac{I \, dL \times \hat{r}}{r^2} \]
- Here, \( dL \) is the length of the current element, \( \hat{r} \) is the unit vector pointing from the current element to the point where we are calculating the field, and \( r \) is the distance from the current element to the center of the loop.

3. **Geometry of the Loop**:
- For a circular loop, the distance \( r \) from any point on the loop to the center is constant and equal to \( R \). Thus, we can simplify the equation:
\[ dB = \frac{\mu_0}{4\pi} \frac{I \, dL}{R^2} \]

4. **Integrating Over the Loop**:
- The total magnetic field \( B \) at the center is obtained by integrating \( dB \) around the entire loop:
\[ B = \int dB = \int \frac{\mu_0}{4\pi} \frac{I \, dL}{R^2} \]
- The integral of \( dL \) over the entire loop is the circumference of the loop, which is \( 2\pi R \):
\[ B = \frac{\mu_0}{4\pi} \frac{I}{R^2} \cdot 2\pi R \]

5. **Simplifying the Expression**:
- Now, simplify the expression:
\[ B = \frac{\mu_0 I}{4\pi} \cdot \frac{2\pi}{R} = \frac{\mu_0 I}{2R} \]

6. **Final Result**:
- Therefore, the magnetic field at the center of the current-carrying circular loop is given by:
\[ B = \frac{\mu_0 I}{2R} \]

### Conclusion:
The magnetic field at the center of a current-carrying circular loop of radius \( R \) is \( \frac{\mu_0 I}{2R} \).