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

To determine the magnetic field at the mid-point of a square wire carrying a current \(I\), we use the Biot-Savart Law. A square wire consists of four straight segments. We need to calculate the contribution to the magnetic field from each segment at the center of the square and sum them up.

The Biot-Savart Law states:

\(d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{l} \times \mathbf{r}}{r^3}\)

Where:

• \(d\mathbf{B}\) is the infinitesimal magnetic field contribution from an infinitesimal length \(d\mathbf{l}\)
• \(\mu_0\) is the permeability of free space
• \(I\) is the current through the wire
• \(\mathbf{r}\) is the position vector from the current element to the point of interest
• \(r\) is the magnitude of \(\mathbf{r}\)

The square's center is at a distance of \(\frac{l}{\sqrt{2}}\) from each vertex. The magnetic field contribution from each side of the square can be calculated independently and is directed along the perpendicular bisector of the side due to symmetry.

The length of each side of the square is \(l\), and the square is symmetrical, so the magnetic fields from opposite sides add up. Consider one side of the square:

The total magnetic field \(\mathbf{B}\) at the center from a side is:

\(\mathbf{B}_{\text{side}} = \frac{\mu_0 I}{\pi l} \sin\theta\)

\(\theta\) for each side is \(45^\circ\), thus \(\sin 45^\circ = \frac{1}{\sqrt{2}}\).

The contribution of magnetic field at the center is:

\(\mathbf{B}_{\text{side}} = \frac{\mu_0 I}{\pi l} \times \frac{1}{\sqrt{2}}\) from one side, and the total is:

\(\mathbf{B}_{\text{total}} = 4 \times \mathbf{B}_{\text{side}}\)

\(\mathbf{B}_{\text{total}} = 4 \times \frac{\mu_0 I}{\pi l} \times \frac{1}{\sqrt{2}}\)

Simplifying gives the final magnitude:

\(\mathbf{B}_{\text{total}} = \frac{2\sqrt{2}\mu_0 I}{\pi l}\)

The correct answer from the options provided is: \(\frac{B}{\sqrt{2}}\).

Answer 2

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: \[ \text{(2) } \frac{B}{\sqrt{2}} \]