9.9 Kq,4 Ar VEs 3.(35 pts) Three very long wires oriented perpendicular to the p
ID: 2039013 • Letter: 9
Question
9.9 Kq,4 Ar VEs 3.(35 pts) Three very long wires oriented perpendicular to the page are each separated by 1 5 sm- current in the left wire is 4.6 A, in the middle wire 8.1 A and in the right wire below the middle wire. A) Find the magnitude and direction the direction as an angle clockwise or counter clockwise from the vertical. B) What the net magnetic force on the far left wire from the other two w 2.7A. The dot is 15 cm of the net magnetic field at the dot. Specify is the direction of ires? Explain.Explanation / Answer
A) let I1 = 4.6 A
I2 = 8.1 A
I3 = 2.7 A
at the given point,
B1 = mue*I1/(2*pi*r1)
= 4*pi*10^-7*4.6/(2*pi*sqrt(2)*0.15)
= 4.34*10^-6 T (45 degrees below -x axis)
B2 = mue*I2/(2*pi*r2)
= 4*pi*10^-7*8.1/(2*pi*0.15)
= 1.08*10^-5 T (towards +x axis)
B3 = mue*I3/(2*pi*r3)
= 4*pi*10^-7*2.7/(2*pi*sqrt(2)*0.15)
= 2.54*10^-6 T (45 degrees below +x axis)
Bnetx = B1x + B2x + B3x
= -4.34*10^-6*cos(45) + 1.08*10^-5 + 2.54*10^-6*cos(45)
= 9.53*10^-6 T
Bnety = B1y + B2y + B3y
= -4.34*10^-6*sin(45) + 0 - 2.54*10^-6*sin(45)
= -4.86*10^-6 T
Bnet = sqrt(Bnetx^2 + Bnety^2)
= sqrt(9.53^2 + 4.86^2)*10^-6
= 1.07*10^-5 T <<<<<<<<<----------------Answer
direction : theta = tan^-1(vx/vy)
= tan^-1(9.53/4.86)
= 63.0 degrees clockwise
B) towards left
because, if current flows in the same direction the wires will attract and
if the current flows in the opposite direction the wires repel.
in the given case the left wire is repeled by the other two wires.
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