Consider the two wires in the figure, running at a right angle to each other. The wires carry the same current I = 2.5 mA, in the directions shown. The distances shown are d1 = 34 cm and d2 = 12.5 cm.A)What is the magnitude of the magnetic field, in tesla, at point A?B) What is the magnitude of the magnetic field, in tesla, at point B? c)What is the magnitude of the magnetic field, in tesla, at point C? The image features a graphical diagram consisting of two perpendicular lines forming a coordinate axis labeled with "I" at both ends of each line. There are three points marked as A, B, and C, each situated along the lines. Here is a detailed description:- **Axes**: The diagram shows a vertical and a horizontal line intersecting at the origin, creating a coordinate plane.- **Points**: - Point **A** is positioned along a horizontal line at a distance \( d_2 \) from the vertical axis. - Point **B** is positioned further to the right along the horizontal line at a distance \( d_1 \) from Point A, and vertically at a distance \( d_1 \) above the horizontal axis. - Point **C** is located below the horizontal axis at a distance \( d_1 \) vertically from the axis and horizontally \( d_2 \) from the vertical axis.This diagram illustrates the distances between points A, B, and C, with specific variables \( d_1 \) and \( d_2 \) representing those distances on the axes.

Given data: Current in the wires, I=2.5 mA = 2.5×10-3 A d1=34 cm = 0.34 m d2=12.5 cm = 0.125 m (a) Magnetic field at point A due to vertical wire: B1=μoI2πd2 Into the plane Magnetic field at point A due to horizontal wire: B2=μoI2πd1 out of the plane Net magnetic field at point A: BA=B1-B2 =μoI2πd2-μoI2πd1 =μoI2π1d2-1d1 Plugging in the given values: BA=4π×10-7×2.5×10-32π10.125-10.34 =25.29×10-10 T (b) Net magnetic field at point B: BB=μoI2πd1-μoI2πd1 =0 (c) Net magnetic field at point C: BC=μoI2πd2+μoI2πd1 =μoI2π1d2+1d1 Plugging in the given values: BC=4π×10-7×2.5×10-32π10.125+10.34 =54.70×10-10 T Answer: (a) BA=25.29×10-10 T (b) BB=0 T (c) BC=54.70×10-10 T (a) Magnetic field at point A due to vertical wire: B1=μoI2πd2 Into the plane Magnetic field at point A due to horizontal wire: B2=μoI2πd1 out of the plane Net magnetic field at point A: BA=B1-B2 =μoI2πd2-μoI2πd1 =μoI2π1d2-1d1 Plugging in the given values: BA=4π×10-7×2.5×10-32π10.125-10.34 =25.29×10-10 T (b) Net magnetic field at point B: BB=μoI2πd1-μoI2πd1 =0 (c) Net magnetic field at point C: BC=μoI2πd2+μoI2πd1 =μoI2π1d2+1d1 Plugging in the given values: BC=4π×10-7×2.5×10-32π10.125+10.34 =54.70×10-10 T Answer: (a) BA=25.29×10-10 T (b) BB=0 T (c) BC=54.70×10-10 T (b) Net magnetic field at point B: BB=μoI2πd1-μoI2πd1 =0 (c) Net magnetic field at point C: BC=μoI2πd2+μoI2πd1 =μoI2π1d2+1d1 Plugging in the given values: BC=4π×10-7×2.5×10-32π10.125+10.34 =54.70×10-10 T Answer: (a) BA=25.29×10-10 T (b) BB=0 T (c) BC=54.70×10-10 T (c) Net magnetic field at point C: BC=μoI2πd2+μoI2πd1 =μoI2π1d2+1d1 Plugging in the given values: BC=4π×10-7×2.5×10-32π10.125+10.34 =54.70×10-10 T Answer: (a) BA=25.29×10-10 T (b) BB=0 T (c) BC=54.70×10-10 T Answer: (a) BA=25.29×10-10 T (b) BB=0 T (c) BC=54.70×10-10 T