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 1: Understand the Setup
We have a circular loop with a radius \( R \) and a current \( I \) flowing through it. We need to find the magnetic field at the center of this loop.

### Step 2: Consider a Small Element
Take a small element \( dl \) on the circular loop. The magnetic field \( dB \) at the center due to this small element can be calculated using the Biot-Savart Law.

### Step 3: Apply the Biot-Savart Law
According to the Biot-Savart Law, the magnetic field \( dB \) at a point due to a current element is given by:
\[ dB = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl \cdot \sin(\theta)}{r^2} \]
Where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current,
- \( dl \) is the length of the current element,
- \( r \) is the distance from the current element to the point where the field is being calculated,
- \( \theta \) is the angle between the current element and the line connecting the element to the point.

### Step 4: Determine the Parameters
In our case:
- The distance \( r \) from the current element \( dl \) to the center of the loop is equal to the radius \( R \) of the loop.
- The angle \( \theta \) is \( 90^\circ \) (since the magnetic field at the center is perpendicular to the plane of the loop), thus \( \sin(90^\circ) = 1 \).

### Step 5: Substitute Values
Substituting these values into the equation:
\[ dB = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl}{R^2} \]

### Step 6: Integrate Over the Loop
To find the total magnetic field \( B \) at the center, we need to integrate \( dB \) around the entire loop:
\[ B = \int dB = \int \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl}{R^2} \]
Since \( R \) and \( I \) are constants, we can take them out of the integral:
\[ B = \frac{\mu_0 I}{4\pi R^2} \int dl \]
The integral \( \int dl \) over the entire loop gives us the circumference of the loop, which is \( 2\pi R \):
\[ B = \frac{\mu_0 I}{4\pi R^2} \cdot 2\pi R \]

### Step 7: Simplify the Expression
Now simplifying the expression:
\[ B = \frac{\mu_0 I}{4\pi R^2} \cdot 2\pi R = \frac{\mu_0 I}{2R} \]

### Final Result
Thus, the magnetic field at the center of the current-carrying circular loop is:
\[ B = \frac{\mu_0 I}{2R} \]