Question

A square loop, with sides of length \(L\), carries current \(i\). Find the magnitude of the magnetic field from the loop at the center of the loop, as a function of \(i\) and \(L\).

Short answer

Based on the solution, the magnitude of the magnetic field at the center of a square loop with side length \(L\) and carrying current \(i\) is given by: \(B = \frac{\mu_0 i L}{\pi}\), where \(\mu_0\) is the permeability of free space.

Step-by-step solution

Biot-Savart Law

Recall the Biot-Savart Law, which states that the small magnetic field d\vec{B} produced by the electric current flowing through a small element d\vec{l} of a conductor at a point P is given by: d\vec{B} = \frac{\mu_0}{4\pi} \frac{i d\vec{l} \times \vec{r}}{r^3} Here, \(\mu_0\) is the permeability of free space, \(i\) is the current through the conductor, and \(\vec{r}\) is the vector from the current element to point P.

Figure out the geometry of the problem

We have a square loop of side \(L\) with current \(i\). Let's assume the loop lies in the xy-plane, with one vertex at the origin, and the loop's sides parallel to the x and y axes. The center of the loop is at \((L/2, L/2)\). We need to find the magnetic field at this point due to the current flowing through all four sides of the loop.

Calculate the magnetic field contribution from one side

Consider one side of the square loop that starts at the origin and extends to \((L, 0)\). We'll look at the \(x\) component of the magnetic field, \(dBx\), at the center \((L/2, L/2)\) due to an element \(d\vec{l}\) located at \((x,0)\). From the Biot-Savart law, we have: d\vec{B} = \frac{\mu_0}{4\pi} \frac{i d\vec{l} \times \vec{r}}{r^3} dBx = \frac{\mu_0 i}{4\pi} \int_{0}^{L} \frac{-dy}{((L/2)^2 + x^2)^{3/2}} To find the total magnetic field contribution \(Bx\) in the \(x\)-direction due to the entire side, we integrate over the length of the side (from \(0\) to \(L\)): Bx = \frac{\mu_0 i}{4\pi} \left[ \frac{(L/2 - x)(L^2/4 + x^2)^{1/2}}{x} \right]_{0}^{L}

Calculate magnetic field contributions from other sides

Using symmetry, we can see that Bx due to the remaining three sides is zero. The same is true for By. All four sides have the magnetic field components in the y-direction that cancel each other at the center. So, we only need to find the magnetic field components in the z-direction. We can calculate the z-component of the magnetic field for each side and sum them up to determine the total magnetic field at the center of the loop.

Calculate the total magnetic field at the center of the loop

Let's denote the magnetic field contribution from each side (in the z-direction) as Bz1, Bz2, Bz3, and Bz4. Bz1 and Bz3 are equal since they are both opposite and parallel to each other. Similarly, Bz2 and Bz4 are equal. The total magnetic field in the z-direction at the center is the sum of all four components: Bz_total = Bz1 + Bz2 + Bz3 + Bz4 = 2(Bz1 + Bz2) By evaluation, we find that: Bz1 = \frac{\mu_0 i L}{4 \pi} \left(\frac{2}{\pi}\right) Bz2 = \frac{\mu_0 i L}{4 \pi} \left(\frac{2}{\pi}\right) Therefore, Bz_total = 2(Bz1 + Bz2) = 2\left(\frac{\mu_0 i L}{4 \pi}\right) \left(\frac{4}{\pi}\right) = \frac{\mu_0 i L}{\pi} So, the magnitude of the magnetic field at the center of the square loop is: \(B = Bz_{\text{total}} = \frac{\mu_0 i L}{\pi}\)