The figure below shows three parallel wires arranged in a equilateral triangle w
ID: 2040469 • Letter: T
Question
The figure below shows three parallel wires arranged in a equilateral triangle with sides of length 10.0 cm. The wire at the apex carries a current of 5.0 A and points out ofthe page while the two wires at the bottom has currents of 8.0 A and 21.0 A pointing into the page. Find the magnitude of the magnetic field at the point equidistant from the wires, using the rules of vector addition to sum the contributions from each wire. 5.0 A 8.0 A 10.0 cm ?21,0 A 9.26e-5T Submit Answer Incorrect. Tries 2/10 Pre vious Tries What is the direction of the field? Specify the angle fromthe +x-axis going counter-clockwise (see figure on right) Submit Answer Tries 0/10 Post Discussion ai, Send FeedbackExplanation / Answer
let L = 10 cm = 0.1 m
I1 = 5 A, I2 = 8 A, I3 = 21 A
let r is the distance from each corner to center of the triangle.
cos(30) = (L/2)/r
r = (L/2)/cos(30)
= (0.1/2)/cos(30)
= 0.0577 m
B1 = mue*I1/(2*pi*r) = 4*pi*10^-7*5/(2*pi*0.0577) = 1.73*10^-5 T
B2 = mue*I2/(2*pi*r) = 4*pi*10^-7*8/(2*pi*0.0577) = 2.77*10^-5 T
B3 = mue*I3/(2*pi*r) = 4*pi*10^-7*21/(2*pi*0.0577) = 7.28*10^-5 T
Bnetx = B1x + B2*cos(60) + B3*cos(60)
= 1.73*10^-5 + 2.77*10^-5*cos(60) + 7.28*10^-5*cos(60)
= 6.76*10^-5 T
Bnety = B1y + B2*cos(60) + B3*cos(60)
= 0 - 2.77*10^-5*sin(60) + 7.28*10^-5*sin(60)
= 3.91*10^-5 T
Bnet = sqrt(Bnetx^2 + Bnety^2)
= sqrt(6.76^2 + 3.91^2)*10^-5
= 7.81*10^-5 T <<<<<<<<<<<<<<<<<<<<<-------------------Answer
direction : theta = tan^-1(Bnety/Bnetx)
= tan^-1(3.91/6.76)
= 30.0 degrees <<<<<<<<<<<<<<<<<<<<<-------------------Answer