91Ó°ÊÓ

The Biot-Savart Law states that the magnetic field strength dB at a point P due to a small segment of wire carrying a current I is given by: \(dB = \frac{\mu_0}{4 \pi} \frac{Id\textbf{l} \times \textbf{r}}{|\textbf{r}|^3}\) Here, \(d\textbf{l}\) is the vector representing the length of a small segment of the wire \(\textbf{r}\) is the position vector from the segment to point P \(I\) is the current in the wire and \(\mu_0 = 4 \pi \times 10^{-7} \mathrm{T \cdot m/A}\) is the magnetic permeability of free space. For our problem, we will find the magnetic field at the center of the loop using the Biot-Savart Law. The center of the loop is equidistant from all its sides.

As the loop is a square, there are four segments (sides) that we need to analyze. Let's call the sides A, B, C, and D. At the center of the loop, on sides A and C, the direction of d\(\textbf{l}\) and \(\textbf{r}\) will be the same. The cross product in the Biot-Savart Law will be zero, and so the magnetic fields produced by sides A and C will cancel each other. For sides B and D, the cross product d\(\textbf{l} \times \textbf{r}\) will be maximal, and the magnetic fields produced by these sides will contribute to the net magnetic field at the center.

For sides B and D, the angle between the position vector \(\textbf{r}\) and the segment vector d\(\textbf{l}\) is 90 degrees, and the position vector has a length equal to half of the loop's side length. Thus, the cross product simplifies as follows: \(d\textbf{l} \times \textbf{r} = d\textbf{l} \cdot |\textbf{r}| = 0.5 \cdot d\textbf{l}\) Now, we can rewrite the Biot-Savart formula for one side: \(dB = \frac{\mu_0}{4\pi} \frac{I(0.5 \cdot d\textbf{l})}{|\textbf{r}|^3}\) For a full side, integrate over the entire length (10.0 cm): \(B_{side} = \int dB = \int^{\text{length}}_0 \frac{\mu_0}{4\pi} \frac{I(0.5 \cdot d\textbf{l})}{|\textbf{r}|^3}\) Finally, consider that the magnetic field produced by sides B and D has the same magnitude and direction. Thus, the net magnetic field is twice the value of the magnetic field produced by one side: \(B_{net} = 2 \cdot B_{side}\)

After substituting values, solving the integral, and finding the net magnetic field, we get: \(B_{net} = 2 \cdot \int_0^{0.1} \frac{4 \pi \times 10^{-7} \mathrm{T \cdot m/A} \cdot 0.300 \mathrm{A} (0.5 \cdot d\textbf{l})}{(0.05 \mathrm{m})^3}\) \(B_{net} = 2 \cdot (1.2 \times 10^{-5} \mathrm{T} \cdot 0.1 \mathrm{m})\) \(B_{net} = 2.4 \times 10^{-6} \mathrm{T}\) So, the magnetic field at the center of the square loop is \(2.4 \times 10^{-6} \mathrm{T}\).