Class Management | Help Practice Problems 9 Magnetic Force, Torque Begin Date: 3
ID: 1787398 • Letter: C
Question
Class Management | Help Practice Problems 9 Magnetic Force, Torque Begin Date: 3/31/2016 12:00:00 AM -- Due Date: 12/8/2017 12:00:00 AM End Date: 12/8/2017 12:00:00 AM (10%) Problem 5: The figure shows a cross section of three long straight current- 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 on 5.00 A -10.0 cm - 10.0 cm- 10.0 A 10.0 cm| 20.0 A Otheexpertta.com DA 17% Part (a) Find the magnitude of the force per units length, in newtons per meter, on wire A. FA = Grade Summary Deductions 0% Potential 100% sin() HOME cos) tan() cotan) asin) acos(0 atan() acotan() sinh) cosh() tanh) | cotanh( ) O Degrees C Radians a ( 7 8 9 AL 4 5 6 * 1 2 3 0 VO BACKSPACE Submissions Attempts remaining: 3 (4% per attempt) detailed view CLEAR Subinit Hint Teedback I give up! Ilints: 3% deduction per hint. Hints reinaiming: 2 Feedback: 5% deduction per feedback. 17% Part (b) Find the magnitude of the force per units length, in newtons per meter, on wire B. E A 17% Part (e) 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 A in 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. DA 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
F/L = o I1 I2 / 2 pi r
o = 4 pi x 10^-7
(a) F/L on wire A
(F/L)B = (4 x pi x 10^-7 x 10 x 5) / (2 x pi x .10) = 0.0001 N/m
(F/L)C = (4 x pi x 10^-7 x 20 x 5) / (2 x pi x .10) = 0.0002 N/m
Both forces are at 60o angle (equilateral triangle)
Cos 60 degrees = .5
(F/L)resultant = sqrt ((.0001)^2 + (.0002)^2 + (2 x .0001 x .0002 x .5))
(F/L)resultant = 2.64 x 10^-4 N/m
(b) F/L on B
(F/L)A = (4 x pi x 10^-7 x 10 x 5) / (2 x pi x .10) = 0.0001 N/m
(F/L)C = (4 x pi x 10^-7 x 20 x 10) / (2 x pi x .10) = 0.0004 N/m
Both forces are at 60o angle (equilateral triangle)
Cos 60 degrees = .5
(F/L)resultant = sqrt ((.0001)^2 + (.0004)^2 + (2 x .0001 x .0004 x .5))
(F/L)resultant = 4.58 x 10^-4 N/m
(c) F/L on C
(F/L)A = (4 x pi x 10^-7 x 5 x 20) / (2 x pi x .10) = 0.0002 N/m
(F/L)B = (4 x pi x 10^-7 x 20 x 10) / (2 x pi x .10) = 0.0004 N/m
Both forces are at 60o angle (equilateral triangle)
Cos 60 degrees = .5
(F/L)resultant = sqrt ((.0002)^2 + (.0004)^2 + (2 x .0002 x .0004 x .5))
(F/L)resultant = 5.29 x 10^-4 N/m
(d) Fres = sqrt (A^2 + B^2 + 2AB Cos (theta)
tan (alpha) = (B sin theta) / (A + B cos theta)
theta = 60 degree
sin 60 = 0.866
Cos 60 = 0.5
tan alpha = (.0002 x .866) / (.0001 + (.0002 x .5)) = .866
alpha = 40.89 degree
From +x = 40.89 + 60 = 100.89 degree