Question

A square wire of each side $ l $ carries a current $ I $. The magnetic field at the mid-point of the square

• A. \( \frac{A}{\sqrt{2}} \)
• B. \( \frac{B}{\sqrt{2}} \)
• C. \( \frac{C}{\sqrt{2}} \)
• D. \( \frac{D}{\sqrt{2}} \)

Hint:

When dealing with a current-carrying square wire, the magnetic field at the center is the vector sum of the fields due to each side, taking into account symmetry.

Answer 1

The Correct Option is B

In this question, we need to find the magnetic field at the center of the square due to the current-carrying sides. 
The magnetic field at the center due to a straight current-carrying wire is given by: \[ B = \frac{\mu_0 I}{4 \pi r} \cdot \text{(magnetic field from one wire)} \] 
However, for a square configuration, each of the four sides of the square will contribute to the magnetic field at the center. 
The total magnetic field at the center is the vector sum of the contributions from all four sides. 
The distance from the center to the mid-point of each side is \( \frac{l}{\sqrt{2}} \), where \( l \) is the side length of the square. 
Therefore, the magnetic field contribution from each side is: \[ B_{\text{total}} = 4 \times \frac{\mu_0 I}{4 \pi \frac{l}{\sqrt{2}}} = \frac{\mu_0 I}{\pi l} \times \sqrt{2} \] 
Thus, the net magnetic field at the center is: \[ B_{\text{total}} = \frac{B}{\sqrt{2}} \] 
So, the correct answer is: \[ \text{(2) } \frac{B}{\sqrt{2}} \]