The magnetic field at the point of intersection of diagonals of a square wire loop of side L, carrying a current I is

Options

A. `(mu_(0)I)/(piL)`
B. `(2mu_(0)I)/(piL)`
C. `(sqrt2mu_(0)I)/(piL)`
D. `(2sqrt2mu_(0)I)/(piL)`

Answer

To find the magnetic field at the point of intersection of the diagonals of a square wire loop of side length \( L \) carrying a current \( I \), we can follow these steps:

### Step 1: Understanding the Geometry
Consider a square wire loop with side length \( L \). The diagonals of the square intersect at the center of the square. The distance from the center to any side of the square is \( \frac{L}{2} \).

### Step 2: Magnetic Field Contribution from One Side
The magnetic field \( B \) at a distance \( r \) from a straight current-carrying wire is given by the formula:
\[ B = \frac{\mu_0 I}{4\pi r} (\sin \alpha + \sin \beta) \]
where \( \alpha \) and \( \beta \) are the angles subtended by the wire at the point where the magnetic field is being calculated.

### Step 3: Determine Angles
For a square loop, when considering the center, the angles \( \alpha \) and \( \beta \) for each side of the square are both \( 45^\circ \). Thus:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Therefore, the contribution to the magnetic field from one side of the square is:
\[ B_1 = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\sin 45^\circ + \sin 45^\circ\right) = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) \]
This simplifies to:
\[ B_1 = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{2}{\sqrt{2}}\right) = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \cdot \sqrt{2} \]

### Step 4: Calculate the Magnetic Field Contribution
Substituting \( r = \frac{L}{2} \):
\[ B_1 = \frac{\mu_0 I \sqrt{2}}{4\pi \left(\frac{L}{2}\right)} = \frac{\mu_0 I \sqrt{2}}{2\pi L} \]

### Step 5: Total Magnetic Field Contribution
Since there are four sides of the square loop, and each contributes equally to the magnetic field at the center, the total magnetic field \( B_{\text{net}} \) is:
\[ B_{\text{net}} = 4 \times B_1 = 4 \times \frac{\mu_0 I \sqrt{2}}{2\pi L} = \frac{2\sqrt{2} \mu_0 I}{\pi L} \]

### Conclusion
Thus, the magnetic field at the point of intersection of the diagonals of the square wire loop is:
\[ B_{\text{net}} = \frac{2\sqrt{2} \mu_0 I}{\pi L} \]

### Final Answer
The magnetic field at the point of intersection of diagonals of a square wire loop of side \( L \), carrying a current \( I \) is:
\[ \frac{2\sqrt{2} \mu_0 I}{\pi L} \]