(b) Find the direction of net magnetic field at \( P_1 \).- ○ Into the page- ○ Right- ○ Out of page- ○ Down- ○ Left- ○ Up **Title: Magnetic Field Calculation for Two Parallel Wires****Problem Statement:**(a) The two wires shown in the figure are separated by a distance \(d\) and carry currents of \(I\) in opposite directions. Find the magnitude of the magnetic field at point \(P_1\).**Diagram Description:**The diagram illustrates two vertical parallel wires separated by a horizontal distance \(d\). The wire on the left carries an upward current \(I\), and the wire on the right carries a downward current \(I\). Point \(P_1\) is located to the right of the second wire, also at a distance \(d\).**Options for the Magnetic Field Magnitude at \(P_1\):**- \(\frac{2\mu_0 I}{\pi d}\)- \(\frac{\mu_0 I}{4\pi d}\)- \(\frac{5\mu_0 I}{2\pi d}\)- \(\frac{3\mu_0 I}{4\pi d}\)- \(\frac{\mu_0 I}{2\pi d}\)In this exercise, students are asked to apply their knowledge of electromagnetism to determine the magnitude of the magnetic field at a specified point due to currents flowing in opposite directions in parallel wires.

Given:Let the wire to the left be 1 and to the right be 2.I1 = +I j^I2 = -I j^Distance between P1 and I1, d1 = 2dDistance between P1 and I2, I2 = dAsked:a. Magnitude of Magnetic field at P1.b. Its direction at P1 Magnetic field due to current I at a distance r isB = μoI2πrand the direction is given in the figure above.Therefore Magnetic field at P1 due to I1 is B1→ = μoI2π(2d)(-k^)B1→ = -μoI4πd(k^)Similarly B2→ =μoI2π(d)(+k^) B2→ = μoI2πd(k^)Therefore net B→= B1→ + B2→B→ = -μoI4πd(k^) + μoI2πd(k^)B→ = +μoI4πd(k^)Therefore magnitude is B = μoI4πd Tesladirection is + k^ that is positive z direction, that is out of the page.

Answered: (a) The two wires shown in the figure…

(a) The two wires shown in the figure are separated by a distance d and carry currents of I in opposite directions. Find the magnitude of the magnetic field at point P1.