A square loop of side a carris a current `I`. The magnetic field at the centre of the loop is

Options

A. `(2mu_0Isqrt2)/(pia)`
B. `(mu_0Isqrt2)/(pia)`
C. `(4mu_0Isqrt2)/(pia)`
D. `(mu_0I)/(pia)`

Explanation

To find the magnetic field at the center of a square loop of side \( a \) carrying a current \( I \), we can follow these steps:

### Step 1: Understand the Geometry of the Square Loop
The square loop has four sides, and we will denote the corners of the square as \( A, B, C, D \). The center of the square, where we want to find the magnetic field, is point \( P \).

### Step 2: Magnetic Field Contribution from Each Side
The magnetic field at the center due to each straight segment of the wire can be calculated using the Biot-Savart Law. For a straight current-carrying wire, the magnetic field \( B \) at a distance \( r \) from the wire is given by:
\[ B = \frac{\mu_0 I}{4 \pi r} \sin \theta \]
where \( \theta \) is the angle between the wire and the line connecting the wire to the point where the field is being calculated.

### Step 3: Determine the Distance and Angles
For each side of the square, the distance from the center \( P \) to the midpoint of each side is \( \frac{a}{2} \). The angle \( \theta \) for each side is \( 45^\circ \) since the current flows perpendicular to the line connecting the center to the midpoint of each side.

### Step 4: Calculate the Magnetic Field from One Side
Using the formula for the magnetic field:
\[ B_{\text{one side}} = \frac{\mu_0 I}{4 \pi \left(\frac{a}{2}\right)} \sin(45^\circ) \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we can substitute this into the equation:
\[ B_{\text{one side}} = \frac{\mu_0 I}{4 \pi \left(\frac{a}{2}\right)} \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 I}{2 \pi a} \cdot \frac{1}{\sqrt{2}} \]

### Step 5: Total Magnetic Field from All Four Sides
Since all four sides contribute equally to the magnetic field at the center, we multiply the contribution from one side by 4:
\[ B_{\text{total}} = 4 \cdot B_{\text{one side}} = 4 \cdot \frac{\mu_0 I}{2 \pi a} \cdot \frac{1}{\sqrt{2}} = \frac{2\sqrt{2} \mu_0 I}{\pi a} \]

### Final Result
Thus, the magnetic field at the center of the square loop is given by:
\[ B = \frac{2\sqrt{2} \mu_0 I}{\pi a} \]