Three parallel wires carrying current I 1 =5.00A, I 2 =4.00A, and I 3 =12.00A ar
ID: 1265301 • Letter: T
Question
Three parallel wires carrying current I1=5.00A, I2=4.00A, and I3=12.00A are configured into an equilateral triangle. The length of one side of the triangle is 0.09m. We want to calculate the force per unit length on I2. What is should you do first?
(a) Calculate the magnitude of the force per unit length between I1 and I2. Well call it F12.
|F12|= N
(b) Calculate the magnitude of the force per unit length between I3 and I2. Well call it F32.
|F32|= N
(c) Calculate the x-components of F12 and F32.
F12x= N
F32x= N
(d) Calculate the y-components of F12 and F32.
F12y= N
F32y= N
(e) Sum the X-components: then sum the y-components.
?Fx= N
?Fy= N
(f) Calculate the resultant vector; magnitude and direction.
|?F|= N
?= o
(counterclockwise from +x-axis)
Explanation / Answer
a) |F12|/L = mue*I1*I2/(2*pi*d)
= 4*pi*10^-7*5*4/(2*pi*0.09)
= 4.44*10^-5 N/m
b) |F32|/L = mue*I3*I2/(2*pi*d)
= 4*pi*10^-7*12*4/(2*pi*0.09)
= 10.67*10^-5 N/m
c) F12x = F12*cos(0) = 4.44*10^-5 N/m
F32x = -F32*cos(60) = -5.33*10^-5 N/m
d) F12y = 0
F32y = F32*sin(60) = 9.24*10^-5 N/m
e) Fnetx = F12x + F32x = -0.89*10^-6 N
Fnety = F12y + F32y = 9.24*10^-5 N/m
f) |Fnet| = sqrt(Fnetx^2 + Fnety^2) = 9.28*10^-5 N/m
direction, theta = tan^-1(Fnety/Fnetx)
= tan^-1(-9.24/0.89)
= 180 - 84.5
= 95.5 degrees with +x axis in counter clockwise direction
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