(6%) Problem 14: The figure shows a cross section of three long straight current

Question

(6%) Problem 14: The figure shows a cross section of three long straight current carrying wires. Noticc we arc using the common convention that crosses denote vectors going into thc pagc and dots denote vectors coming out of the page 5.00 A 10.0 A 1 10.0 cm-+1 20.0 A ©theexpertta.com 17% Part (a) Find the magnitude of the force per units length, in newtons per meter, on wire A Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: S cos sin cotan) asinO atan)acotan)sinh0 cOS acos() % per attempt) detailed view cosh0 tanh0 cotan END Degrees Radians BACKSPACE DFLCLEAR Submit Hint I give up! Hints:-o-deduc tion per hint. Hints remaining: 2 Feedback: 0% deduction per feedback Submission Historv Answer Totals 0% Hints Feedbaclk Totals 0% 0% 0% 17% Part (b) Find the magnitude of the force per units length, in newtons per meter, on wire B 17% Part (c) Find the magnitude of the force per units length, in newtons per meter, on wire C 17% Part (d) Find the direction of the force for wire A in degrees. Measure this angle counterclockwise in degrees, from the positive x- axis 17% Part (e) Find the direction of the force for wire B, in degrees. Measure this angle counterclockwise, in degrees, from the positive x- axis 17% Part (f) Find the direction of the force for wire C in degrees. Measure this angle counterclockwise in degrees, from the positive x- axis

Explanation / Answer

(a)

FBAx = uo*Ib*Ia*cos60/(2*pi*r) = 4pi*10^-7*10*5*cos60/(2*pi*0.1) = 5*10^-5 N/m

FBAy = uo*Ib*Ia*sin60/(2*pi*r) = 4pi*10^-7*10*5*sin60/(2*pi*0.1) = 8.66*10^-5 N/m

FCAx = -uo*Ic*Ia*cos60/(2*pi*r) = -4pi*10^-7*20*5*cos60/(2*pi*0.1) = -10*10^-5 N/m

FCAy = uo*Ic*Ia*sin60/(2*pi*r) = 4pi*10^-7*20*5*sin60/(2*pi*0.1) = 17.32*10^-5 N/m

FAx = FBAx + FCAx = -5*10^-5 N/m

FAy = FBAy + FCAy = 25.98*10^-5 N/m

FA = sqrt(FAX^2 + FAy^2) = 26.45*10^-5 N/m

(b)

FABx = -uo*Ib*Ia*cos60/(2*pi*r) = -4pi*10^-7*10*5*cos60/(2*pi*0.1) = -5*10^-5 N/m

FBAy = -uo*Ib*Ia*sin60/(2*pi*r) = -4pi*10^-7*10*5*sin60/(2*pi*0.1) = -8.66*10^-5 N/m

FCBx = uo*Ic*Ib*cos0/(2*pi*r) = 4pi*10^-7*20*10*cos0/(2*pi*0.1) = 40*10^-5 N/m

FCBy = 0

FBx = FABx + FCBx = 35*10^-5 N/m

FBy = FABy + FCBy = -8.66*10^-5 N/m

FB = sqrt(FBX^2 + FBy^2) = 36.05*10^-5 N/m

(c)

FACx = uo*Ic*Ia*cos60/(2*pi*r) = 4pi*10^-7*20*5*cos60/(2*pi*0.1) = 10*10^-5 N/m

FACy = -uo*Ic*Ia*sin60/(2*pi*r) = -4pi*10^-7*20*5*sin60/(2*pi*0.1) = -17.32*10^-5 N/m

FBCx = -uo*Ib*Ic*cos0/(2*pi*r) = -4pi*10^-7*10*20*cos0/(2*pi*0.1) = -40*10^-5 N/m

FBCy = 0

FCx = FACx + FBCx = -30*10^-5 N/m

FCy = FACy + FBCy = -17.32*10^-5 N/m

Fc = sqrt(FcX^2 + Fcy^2) = 34.6*10^-5 N/m

(d)

tan^-1(FAy/FAx) = 101 degrees

(e)

tan^-1(FBy/FBx) = 346 degrees

(f)

tan^-1(Fcy/Fcx) = 210 degrees

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