carrying wires. Notice we are using the common convention that crosses denote ve
ID: 1599278 • Letter: C
Question
carrying wires. Notice we are using the common convention that crosses denote vectors going into the page and dots denote vectors coming out of the page. 5.00 A C 10.0 A H 10.0 cm 20.0 A Otheexpertta com A 17% Part (a) Find the magnitude of the force per units length, in newtons per meter, on wire A. Grade Summary Deductions Potential 100% sino cos0 tano 7 8 9 Submissions HOME Attempts remaining: 8 cotano asino acos0 4 5 6 (09 per attempt) detailed view atan0 acotan0 sinh0 1 2 3 cosh0 tanh0 cotanh0 vo Degrees Radians BACKSPACE Submit Hint I give up! Hints 0% deduction per hint. Hints remaining 2 Feedback 0% deduction per feedback. A 17% Part (b) Find the magnitude of the force per units length, in newtons per meter, on wire B A 17% Part (c Find the magnitude of the force per units length, in newtons per meter, on wire C A 17% Part (d Find the direction of the force for wire Ain degrees. Measure this angle counterclockwise, in degrees, from the positive x. axis A 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. A 17% Part (f Find the direction of the force for wire C, in degrees. Measure this angle counterclockwise, in degrees, from the positive x- axisExplanation / Answer
Fab = muo*I1*I2/ 2*pi*d
Fab is force per unit length on wire A due to wire B
Fab = 2*10^-7*10*5/10 = 10*10^-6 N
Fac = muo*I1*I2/ 2*pi*d
Fac is force per unit length on wire A due to wire C
Fac = 2*10^-7*5*20/10 = 2*10^-5 N
Net Force per unit length = sqrt( Fab^2 + Fac^2 + 2Fab*Fac*Cos60) = 2.64*10^-5 N
B.)
Fba = muo*I1*I2/ 2*pi*d
Fba is force per unit length on wire B due to wire A
Fba = 2*10^-7*10*5/10 = 10*10^-6 N
Fbc = muo*I1*I2/ 2*pi*d
Fbc is force per unit length on wire B due to wire C
Fbc = 2*10^-7*10*20/10 = 4*10^-5 N
Net Force per unit length = sqrt( Fba^2 + Fbc^2 + 2Fba*Fbc*Cos60) = 5*10^-5 N
C)
Fca = muo*I1*I2/ 2*pi*d
Fca is force per unit length on wire C due to wire A
Fca = 2*10^-7*20*5/10 = 2*10^-5 N
Fcb = muo*I1*I2/ 2*pi*d
Fcb is force per unit length on wire C due to wire B
Fcb = 2*10^-7*5*20/10 = 2*10^-5 N
Net Force per unit length = sqrt( Fca^2 + Fcb^2 + 2Fca*Fcb*Cos60) = 4*10^-5 N
D.)Direction of Resultant Vector = tan-1 [(2x10^-5 x sin 60°) / (10^-5 + 2x10^-5 x cos 60°)]
= 49.10 degree
angle from x - axis = 60+49.1 = 109.1 degree
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