Two straight parallel wires carry currents in opposite directions as shown in th

Question

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of 12 = 11.7 A. Point A is the midpoint between the wires. The total distance between the wires is d = 13.0 cm. Point C is 4.72 cm to the right of the wire carrying current I2. Current I1 is adjusted so that the magnetic field at C is zero. Calculate the value of the current I1. Calculate the magnitude of the magnetic field at point A. What is the force between two 1.83 m long segments of the wires?

Explanation / Answer

from ampere's law

integral B.dL = mu*i

B = mu*i/2pid

magnetic field at C is zero

BI1-BI2 = 0

BI1 = BI2

muI1/2pi*(13+4.72)*10^-2 = muI2/2pi*4.72x10^-2

I1/17.72 = 11.7/4,72

I1 = 43.92 A

part b)

BI1 = muI1/2pid

here d = 13/2 = 6.5 cm = 6.5 x 10^-2 m

same for BI2

BI2 = muI2/2pid

Bnet = BI1 + BI2

because Using the Right Hand Rule, we see that these two fields point in the same direction -- "up" or "out of the page" at point A

Bnet = muo/2pid * (I1 + I2)

Bnet = 1.71 x 10^-4

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.