Suppose in the figure that the four identical currents i = 14 A , into or out of
ID: 2260238 • Letter: S
Question
Suppose in the figure that the four identical currents i = 14 A, into or out of the page as shown, form a square of side 150 cm. What is the force per unit length (magnitude and direction) on the wire in the bottom left hand corner? (N/m) Take the positive y direction as up and the positive x direction as to the right.
1. What is the magnitude of the force per unit length?
2. What is the x component of the force per unit length?
3. What is the y component of the force per unit length?
Suppose in the figure that the four identical currents i = 14 A, into or out of the page as shown, form a square of side 150 cm. What is the force per unit length (magnitude and direction) on the wire in the bottom left hand corner? (N/m) Take the positive y direction as up and the positive x direction as to the right. What is the magnitude of the force per unit length? What is the x component of the force per unit length? What is the y component of the force per unit length?Explanation / Answer
F1 = u*i^2/(2 pi a) = 4*3.14e-7*14*14/(2*3.14*1.50) = 2.6133 x 10^-5 N/m
F1y = 2.6133 x 10^-5 N/m
F1x = 0
F2 = u*i^2/(2 pi a) = 4*3.14e-7*14*14/(2*3.14*1.50) = 2.6133 x 10^-5 N/m
F2x = -2.6133 x 10^-5 N/m
F2y = 0
F3 = u*i^2/(2 pi sqrt(2) a) = 4*3.14e-7*14*14/(2*3.14*sqrt(2)*1.50) = 1.8479 x 10^-5 N/m
F3x = -(sqrt(2)/2) F3 = -1.3067 x 10^-5 N/m
F3y = -(sqrt(2)/2) F3 = -1.3067 x 10^-5 N/m
Fx = F1x + F2x + F3x = -3.92 x 10^-5 N/m
Fy = F1y+F2y+F3y = 1.3066 x 10^-5 N/m
F = sqrt(3.92*3.92+1.3066*1.3066) = 4.13 x 10^-5 N/m
1. What is the magnitude of the force per unit length?
4.13 x 10^-5 N/m
2. What is the x component of the force per unit length?
-3.92 x 10^-5 N/m
3. What is the y component of the force per unit length?
1.31 x 10^-5 N/m