Three long, parallel conductors each carry a current of I = 2.24 A. The figure b
ID: 1494433 • Letter: T
Question
Three long, parallel conductors each carry a current of I = 2.24 A. The figure below is an end view of the conductors, with each current coming out of the page. Taking a = 1.25 cm, determine the magnitude and direction of the magnetic field at the following points.
Three long, parallel conductors each carry a current of I = 2.24 A. The figure below is an end view of the conductors, with each current coming out of the page. Taking a = 1.25 cm, determine the magnitude and direction of the magnetic field at the following points (a) point A magnitude direction toward the bottom of the page (b) pointlB magnitude direction toward the bottom of the page (c) point C magnitude O direction no direction HTExplanation / Answer
The magnitude of magnetic field is given by
B = u0*I/(2*pi*r)
at point A
distance of point A from two wire = sqrt(2)*a = 1.77 cm = 0.0177 m
distance of point A from third wire = 3*a = 3.75 cm = 0.0375 m
now magnetic field due to one wire at point A
B1 = u0*2.24/(2*pi*0.0177) = (4*pi*10^(-7))*2.24/(2*pi*0.0177) = 25.31*10^(-6) T
magnetic field due to second wire at point A
B2 = u0*2.24/(2*pi*0.0177) = (4*pi*10^(-7))*2.24/(2*pi*0.0177) = 25.31*10^(-6) T
net magnetic field of B1 and B2 = sqrt(B1^2 + B2^2) = 35.79*10^(-6) T and direction is -y axis
magnetic field due to third wire at point A
B3 = u0*2.24/(2*pi*0.0375) = (4*pi*10^(-7))*2.24/(2*pi*0.0375) = 11.94*10^(-6) T and direction is -y axis.
net magnetic field at point A = 35.79*10^(-6) + 11.94*10^(-6) = 47.73*10^(-6) T and direction is -y axis.
at point B
magnetic field due to one wire at point B
B1 = u0*2.24/(2*pi*0.0125) = (4*pi*10^(-7))*2.24/(2*pi*0.0125) = 35.84*10^(-6) T direction is +x axis.
magnetic field due to second wire at point B
B2 = u0*2.24/(2*pi*0.0125) = (4*pi*10^(-7))*2.24/(2*pi*0.0125) = 35.84*10^(-6) T direction is -x axis.
net magnetic field of B1 and B2 = B1 - B2 = 0
magnetic field due to third wire at point B
B3 = u0*2.24/(2*pi*2*0.0125) = (4*pi*10^(-7))*2.24/(2*pi*0.0125) = 35.84*10^(-6) T and direction is -y axis.
net magnetic field at point B
B = B3 = 35.84*10^(-6) T and direction is -y axis.
at point C
magnetic field due to one wire at point C
B1 = u0*2.24/(2*pi*0.0177) = (4*pi*10^(-7))*2.24/(2*pi*0.0177) = 25.31*10^(-6) T
magnetic field due to second wire at point C
B2 = u0*2.24/(2*pi*0.0177) = (4*pi*10^(-7))*2.24/(2*pi*0.0177) = 25.31*10^(-6) T
net magnetic field of B1 and B2 = sqrt(B1^2 + B2^2) = 35.8*10^(-6) T and direction is +y axis
magnetic field due to third wire at point C
B3 = u0*2.24/(2*pi*0.0125) = (4*pi*10^(-7))*2.24/(2*pi*0.0125) = 35.8*10^(-6) T and direction is -y axis.
net magnetic field at point C = 35.8*10^(-6) - B3 = 0
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.