Four identical currents Please show your work and explain how you got your answe

Question

Four identical currents

Please show your work and explain how you got your answers for each part! Thank you!

If you don't know how to do this, please allow someone else to work it out!Suppose in the figure that the four identical currents i = 7 A, into or out of the page as shown, form a square of side 138 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. (Figure Included Here)

A) What is the magnitude of the force per unit length? Check the units for MKS.

B) What is the x component of the force per unit length?

C) What is the y component of the force per unit length?

(8c29p36) Suppose in the figure that the four identical currents i 7 A, into or out of the page as shown, form a square of side 138 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?

Explanation / Answer

(A0

formula for magnetic field due to current carrying wire is

B = uo I/ 2pi r

the magnitude of the force per unit length is

F/L = uo I1 I2/ 2pi d

(a)

F1/L = 4pi * 10^-7 ( 7A) (7A)/ 2pi (1.38 m) = 71* 10^-7 N/m

F2/L = 4pi * 10^-7 ( 7A) (7A)/ sqrt 2 * 2pi (1.38 m) =50.22* 10^-7 N/m

F3/L = 4pi * 10^-7 ( 7A) (7A)/ 2pi (1.38 m) = 71* 10^-7 N/m

Fx = F3 - F2 cos 45 = -71* 10^-7 N/m- 50.22* 10^-7 N/m cos 45 = -106.51* 10^-7 N/m

Fy = F1- F2 sin 45 =  71* 10^-7 N/m- 50.22* 10^-7 N/m cos 45 = 35.48* 10^-7 N/m

F magnitude = sqrt (Fx^2 + Fy^2) = sqrt ( -106.51* 10^-7)^2 + ( 35.48 * 10^-7)^2 = 1.12 * 10^-5 N/m

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