The magnitude of the net magnetic field at the empty corner is 0.396 mT.[tex]________________________________[/tex]To determine the magnitude of the magnetic field at the empty corner of the square, we will use Biot-Savart’s Law or Ampère’s Law, particularly the equation for the magnetic field due to a long, straight current-carrying wire:[tex]\[B = \frac{\mu_0 I}{2 \pi r}\][/tex]where: [tex]\( B \)[/tex] is the magnetic field at distance [tex]\( r \)[/tex], [tex]\( \mu_0 = 4\pi \times 10^{-7} \)[/tex] T·m/A (permeability of free space), [tex]\( I = 99.0 \)[/tex] A (current in each wire), [tex]\( r = 5.0 \)[/tex] cm [tex]\( = 0.050 \)[/tex] m (distance from the wire to the corner). Step 1: Find the Magnetic Field from One Wire For each wire, the magnetic field at the empty corner is:[tex]\[B = \frac{(4\pi \times 10^{-7}) (99.0)}{2\pi (0.050)}\][/tex][tex]\[B = \frac{(4 \times 10^{-7} \times 99)}{(2 \times 0.050)}\][/tex][tex]\[B = \frac{3.96 \times 10^{-5}}{0.1}\][/tex][tex]\[B = 3.96 \times 10^{-4} \text{ T} = 0.396 \text{ mT}\][/tex]Each wire contributes a field of 0.396 mT at the empty corner.Step 2: Find the Net Magnetic Field Since the three wires are at the corners of a square, their magnetic fields form a symmetric pattern at the empty corner. Using the right-hand rule, the fields add up as vectors. By symmetry, the net field is found using vector addition at 120° angles. Using the vector sum formula:[tex]\[B_{\text{net}} = B \sqrt{2 + 2\cos 120^\circ}\][/tex]Since [tex]\(\cos 120^\circ = -\frac{1}{2}\)[/tex], we get:[tex]\[B_{\text{net}} = B \sqrt{2 - 1}\][/tex][tex]\[B_{\text{net}} = B \times \sqrt{1}\][/tex][tex]\[B_{\text{net}} = 0.396 \text{ mT}\][/tex]Thus, the magnitude of the net magnetic field at the empty corner is 0.396 mT.