Question

8. As shown in the figure, two long, parallel wires are separated by 0.50 meters. The current in wire B is 7.0 A upwards and 5.0 A downward in wire A. A 15-turn coil is centered between the two wires and spans from wire B to wire A. What must be the magnitude and direction of the current in the coils for the magnetic field to be zero at the center of the coil? (8 points) +4 7.0 som - 5.0A Brnet wires - Bal=0 B&B -S.) D? (5.6-4) 10-5) = Mo (15) I 1025m=R M. (TA) (4*10-)(7) Zn(,025) 27.025) = 5.6*10-5 ſ, U. (SA) V (47<10-1)() ID 20 (1025) 2+(,025) = 4 x 10-3 Borl = elon I I (1.6x103) I = (478x10-1 (15) T- A7

Answer 1

The field due to the two wires will be into the page and hence they will add together. Thus the field due to coil must be out of page

At the center, which is also the midpoint of separation between wires, the field due to the two wires is into the page 0.25 0.25 В - So, Bar = BA + BB = MoIA + Molb 270.25m) 27(0.25m) IA =5A, IB = 7A in & BAB = 9.6810-6 7 into the page So, field due to coil must be out of page direction and equal to BAB in magnitude So, for the coil, B = Nello I - 28 r=0.5 = 0.25m > 3 = BA = Mo I 7 9.6x10-6 T = NMO I - 2/0.25m) I put N=1s 1 = 70-255) A counter clockwise

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