Final answer:
The magnetic field at the center of the square cross-section of a toroid with the given dimensions and parameters is 4.02 mT.
Step-by-step explanation:
To calculate the magnetic field strength at the center of the square cross-section of a toroid, we use Ampère's Law, which states that the magnetic field (B) around a closed loop is directly proportional to the electric current passing through the loop. The formula for the magnetic field inside a toroid is given by B = \(\frac{\mu_0 N I}{2 \pi r}\), where \(\mu_0\) is the permeability of free space \(\mu_0 = 4\pi \times 10^{-7} T\cdot m/A\), N is the number of turns, I is the current, and r is the radius of the toroid.
In this example, the toroid has a square cross-section of 3.0 cm x 3.0 cm, an inner radius of 25.5 cm, 510 turns of wire, and carries a current of 3.2 A. Plugging in these values into the formula, we get:
B = \(\frac{(4\pi \times 10^{-7} T\cdot m/A) \times 510 \times 3.2 A}{2 \pi \times 0.255 m}\) = \(\frac{2.048 \times 10^{-3} T\cdot m}{0.510 m}\) = 4.02 \times 10^{-3} T
Thus, the magnetic field at the center of the square cross section of the toroid is 4.02 mT (milliteslas).