A transmission wire oriented parallel to the x-axis carries a current of 200 A f
ID: 1774327 • Letter: A
Question
A transmission wire oriented parallel to the x-axis carries a current of 200 A flowing along the +x direction. A second transmission wire also oriented parallel to the x-axis but lying at a perpendicular distance of 1.15 m below the first wire carries a current of 235 A flowing along the +x direction. What is the magnitude of the total magnetic field at a point midway between the two wires? What is the direction of the net magnetic field?
A transmission wire oriented parallel to the x-axis carries a current of 200 A flowing along the +x direction. A second transmission wire also oriented parallel to the x-axis but lying at a perpendicular distance of 1.15 m below the first wire carries a current of 235 A flowing along the +x direction. What is the magnitude of the total magnetic field at a point 2.3 m directly below a point midway between the two wires? What is the direction of the net magnetic field?
A transmission wire oriented parallel to the x-axis carries a current of 200 A flowing along the +x direction. A second transmission wire also oriented parallel to the x-axis but lying at a perpendicular distance of 1.15 m below the first wire carries a current of 235 A flowing along the +x direction. If the location where the net magnetic field is zero is at perpendicular distance X from the first wire, what is X ?
A transmission wire oriented parallel to the x-axis carries a current of 200 A flowing along the +x direction. A second transmission wire also oriented parallel to the x-axis but lying at a perpendicular distance of 1.15 m below the first wire carries a current of 235 A flowing along the -x direction. What is the magnitude of the total magnetic field at a point 2.3 m directly above a point midway between the two wires?
Explanation / Answer
current in first wire I1 = 200 A
current in second wire I2 = 235 A
midpoint is at r = 1.15/2 = 0.575 m
magnetic field due to wire 1 , Bz1 = -uo*I1/(2*pi*r) in to the page
magnetic field due to wire 2 , B2z = uo*I2/(2*pi*r) out of the page
net mangnetic field Bnet = B1 + B2
Bnet = (uo/(2*pi*r))*(-I1 + I2)
Bnet = (4*pi*10^-7/(2*pi*0.575))*(-200+235)
Bnet = 1.22*10^-5 T out of page
magnitude 1.22*10^-5 T
direction = +Z direction
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current in first wire I1 = 200 A
current in second wire I2 = 235 A
distance of point from wire 1, r1 = 1.15/2 + 2.3 = 2.875 m
distance of point from wire 2, r2 = 2.3 - 1.15/2 = 1.725 m
magnetic field due to wire 1 , Bz1 = -uo*I1/(2*pi*r1) in to the page
magnetic field due to wire 2 , B2z = -uo*I2/(2*pi*r2) in to the page
net mangnetic field Bnet = B1 + B2
Bnet = (uo/(2*pi))*(-I1/r1 - I2/r2)
Bnet = (4*pi*10^-7/(2*pi))*(-200/2.875 - 235/1.725)
Bnet = -4.12*10^-5 T into the page
magnitude = 4.12*10^-5 T
direction = Z direction
===================================================
current in first wire I1 = 200 A
current in second wire I2 = 235 A
distance of point X from wire 1, r1 = x
distance of point X from wire 2, r1 = 1.15 - x
magnetic field due to wire 1 , Bz1 = -uo*I1/(2*pi*r1) in to the page
magnetic field due to wire 2 , B2z = uo*I2/(2*pi*r2) out of the page
net mangnetic field Bnet = B1 + B2
Bnet = (uo/(2*pi))*(-I1/r1 + I2/r2)
given Bnet = 0
Bnet = (4*pi*10^-7/(2*pi))*(-200/x + 235/(1.15-x)) = 0
x =0.53 m <<-----ANSWER
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current in first wire I1 = 200 A
current in second wire I2 = 235 A
distance of point from wire 1, r1 = 2.3 - 1.15/2 = 1.725 m
distance of point from wire 2, r2 = 1.15/2 + 2.3 = 2.875 m
magnetic field due to wire 1 , Bz1 = uo*I1/(2*pi*r1) out of the page
magnetic field due to wire 2 , B2z = -uo*I2/(2*pi*r2) in to the page
net mangnetic field Bnet = B1 + B2
Bnet = (uo/(2*pi))*(-I1/r1 - I2/r2)
Bnet = (4*pi*10^-7/(2*pi))*(200/1.725 - 235/2.875)
Bnet = 6.84*10^-6 T out of the page
magnitude = 6.84*10^-6 T
direction = +Z direction
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