A wire in the from of a square of side `'a'` carries a current `i`. Then the magnetic induction at the centre of the square wire is (Magnetic permeability of free space=`mu_(0)`)

Options

A. `(mu_(0)i)/(2pia)`
B. `(mu_(0)isqrt2)/(pia)`
C. `(2sqrt2mu_(0)i)/(pia)`
D. `(mu_(0)i)/(sqrt2 pia)`

Answer

To find the magnetic induction (magnetic field) at the center of a square wire carrying current, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a square wire of side length \( a \) carrying a current \( i \).
- We need to calculate the magnetic field at the center \( O \) of the square.

2. **Magnetic Field due to a Straight Wire**:
- The magnetic field \( B \) at a distance \( R \) from a long straight wire carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi R} \]
- Here, \( \mu_0 \) is the permeability of free space.

3. **Distance from the Center to the Wire**:
- The distance \( R \) from the center \( O \) to the midpoint of each side of the square is \( \frac{a}{2} \).

4. **Calculating the Magnetic Field from One Side**:
- For one side of the square, we can substitute \( R = \frac{a}{2} \) into the magnetic field formula:
\[ B_{\text{one side}} = \frac{\mu_0 i}{2 \pi \left(\frac{a}{2}\right)} = \frac{\mu_0 i}{\pi a} \]

5. **Considering the Contribution from All Four Sides**:
- Since the square has four sides and the magnetic field contributions from each side will have the same magnitude but different directions, we need to consider the angles.
- The angle \( \theta \) between the wire and the line connecting the wire to the center \( O \) is \( 45^\circ \) for each side.

6. **Using the Sine Component**:
- The effective magnetic field contribution from each side at the center is:
\[ B_{\text{effective}} = B_{\text{one side}} \cdot \sin(45^\circ) = \frac{\mu_0 i}{\pi a} \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 i}{\sqrt{2} \pi a} \]

7. **Total Magnetic Field at the Center**:
- Since there are four sides contributing equally, the total magnetic field \( B_{\text{total}} \) at the center is:
\[ B_{\text{total}} = 4 \cdot B_{\text{effective}} = 4 \cdot \frac{\mu_0 i}{\sqrt{2} \pi a} = \frac{4 \mu_0 i}{\sqrt{2} \pi a} = \frac{2 \sqrt{2} \mu_0 i}{\pi a} \]

### Final Result:
The magnetic induction at the center of the square wire is:
\[ B = \frac{2 \sqrt{2} \mu_0 i}{\pi a} \]