incorrect **Educational Resource: Magnetic Fields of Parallel Wires****Figure Description:**The figure illustrates two wires carrying currents of \( I = 4.95 \, \text{A} \) in opposite directions. The wires are separated by a distance \( d \), which is assumed to be 8.0 cm. The wires are labeled, with one on the left and one on the right. There are three points of interest labeled \( P_1 \), \( P_2 \), and a point halfway between the wires, positioned perpendicular to the wires.---**Problem Analysis:****(a)** Find the magnitude (in \(\mu \text{T}\)) and direction of the net magnetic field at a point halfway between the wires.- **Magnitude:** 95 \(\mu \text{T}\) (incorrect calculation)- **Direction:** Into the page (verified correct)**Instruction:** Use the right-hand rule to determine the direction. The vector magnetic fields from each wire at the point should be calculated and combined to find the net field.---**(b)** Find the magnitude (in \(\mu \text{T}\)) and direction of the net magnetic field at point \( P_1 \), 8.0 cm to the right of the wire on the right.- **Magnitude:** 1.2 \(\mu \text{T}\) (incorrect calculation)- **Direction:** Out of the page (verified correct)---**(c)** Find the magnitude (in \(\mu \text{T}\)) and direction of the net magnetic field at point \( P_2 \), \( 2d = 16.0 \) cm to the left of the wire on the left.- **Magnitude:** 4.0 \(\mu \text{T}\) (incorrect calculation)- **Direction:** Out of the page (verified correct)---**(d)** What If? Repeat parts (a) through (c) assuming both wires carry currents in the same direction, upward in the figure.1. **Net Magnetic Field at Halfway Point:** - **Magnitude:** \(0\) \(\mu \text{T}\) (correct) - **Direction:** No direction (fields cancel each other)2. **Net Magnetic Field at \( P_1 \), 8.0 cm to the right of the wire on the right:** - **Magnitude

a) Halfway between the wires (distance d2) the feild due to the left wire is into the page and due to wire on right is also into the page (via right hands thumb rule). Hence the two fields add up. Hence the required magnetic field, B=B1+B2=μ0i2πd/2+μ0i2πd/2=μ0iπd+μ0iπd=2μ0iπd=μ04π8id=10-78·4.950.08=49.5×10-6 T=49.5 μT b) The point P1 is at a distance 2d from the left wire and the magnetic field due to this wire is pointing into the page. This point is at a distance d from the right wire and field due to this sire is pointing out of the page. Hence the net field out of the page is, B=B2-B1=μ0i2πd-μ0i2π2d=μ0i4πd=μ04πid=10-74.950.08=6.1875×10-6=6.1875 μT c) The point P2 is at a distance 2d from the left wire and the magnetic field due to this wire is pointing out of the page. This point is at a distance 3d from the right wire and field due to this sire is pointing into the page. Hence the net field out of the page is, B=B1-B2=μ0i2π2d-μ0i2π3d=μ0i2πd12-13=μ0i2πd16=μ0i4πd26=μ04πi3d=10-74.953·0.08=2.0625×10-6=2.0625 μT As a part of Bartleby guidelines, only this subparts of a questions can be answered at once. Please post for the remaining parts separately.