Three long, parallel conductors each carry a current of I = 2.22 A. The figure b
ID: 1790283 • Letter: T
Question
Three long, parallel conductors each carry a current of I = 2.22 A. The figure below is an end view of the conductors, with each current coming out of the page. Taking a = 1.55 cm, determine the magnitude and direction of the magnetic field at the following points.
Three long, parallel conductors each carry a current of I = 2.22 A. The figure below is an end view of the conductors, with each current coming out of the page. Taking a = 1.55 cm, determine the magnitude and direction of the magnetic field at the following points. aI (a) point A magnitude 1.38e-8 direction Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. toward the bottom of the page (b) point B magnitude Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all direction toward the bottom of the page 14 intermediate results to at least four-digit accuracy to minimize roundoff error. UT (c) point C magnitude 0 direction no direction HT Need Help? Read It Submit Answer Save Progress Practice Another VersionExplanation / Answer
magnitude ofmagnetic field due to current wire B = uo*I/(2*pi*1)
at pointA
x component magnetic field due to top wire B1x = (uo*I/2*pi*a*sqrt2)*cos45
y component magnetic field due to top wire B1y = -(uo*I/2*pi*a*sqrt2)*sin45
x component magnetic field due to bottom wire B2x = -(uo*I/2*pi*a*sqrt2)*cos45
y component magnetic field due to bottom wire B2y = -(uo*I/2*pi*a*sqrt2)*sin45
x component magnetic field due to right wire B3x = 0
y component magnetic field due to right wire B3y = -(uo*I/2*pi*3a)
Bx = B1x + B2x + B3x = 0
By = B1y + B2y + B3y
By = -2*(uo*I/2*pi*a*sqrt2)*sin45 - (uo*I/2*pi*3a)
By = -(uo*I/(2*pi*a))*(2*sin45/sqrt2 + 1/3)
By = -(4*pi*10^-7*2.22/(2*pi*0.0155))*(2*sin45/sqrt2 + 1/3)
By = -3.82*10^-5 T
B = sqrt(Bx^2 + By^2)
direction = tan^-1(By/Bx)
(a)
magnitude = 38.2 uT
direction towards bottom of page
===================================
point B
x component magnetic field due to top wire B1x = (uo*I/2*pi*a)
y component magnetic field due to top wire B1y = 0
x component magnetic field due to bottom wire B2x = -(uo*I/2*pi*a)
y component magnetic field due to bottom wire B2y = 0
x component magnetic field due to right wire B3x = 0
y component magnetic field due to right wire B3y = -(uo*I/2*pi*2a)
Bx = B1x + B2x + B3x = 0
By = B1y + B2y + B3y
By = -(uo*I/2*pi*2a)
By = -(4*pi*10^-7*2.22/(2*pi*2*0.0155))
By = -14.3 uT
B = sqrt(Bx^2 + By^2)
direction = tan^-1(By/Bx)
magnitude = 14.3 uT
direction toward the bottom of page
============================
point C
x component magnetic field due to top wire B1x = (uo*I/2*pi*a*sqrt2)*cos45
y component magnetic field due to top wire B1y = (uo*I/2*pi*a*sqrt2)*sin45
x component magnetic field due to bottom wire B2x = -(uo*I/2*pi*a*sqrt2)*cos45
y component magnetic field due to bottom wire B2y = (uo*I/2*pi*a*sqrt2)*sin45
x component magnetic field due to right wire B3x = 0
y component magnetic field due to right wire B3y = -(uo*I/2*pi*a)
Bx = B1x + B2x + B3x = 0
By = B1y + B2y + B3y
By = 2*(uo*I/2*pi*a*sqrt2)*sin45 - (uo*I/2*pi*a)
By = (uo*I/(2*pi*a))*(2*sin45/sqrt2 - 1)
By = 0
By = 0 T
B = sqrt(Bx^2 + By^2)
direction = tan^-1(By/Bx)
(c)
magnitude = 0 uT
no direction
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.