1 2 3 4 5 6 7 Three infinite straight wires are fixed in place and aligned paral
ID: 777269 • Letter: 1
Question
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Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The wire at (x,y) = (-18.5 cm, 0) carries current I1 = 2 A in the negative z-direction. The wire at (x,y) = (18.5 cm, 0) carries current I2 = 1.3 A in the positive z-direction. The wire at (x,y) = (0, 32 cm) carries current I3 = 6.9 A in the positive z-direction.
1)
What is Bx(0,0), the x-component of the magnetic field produced by these three wires at the origin?T
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2)
What is By(0,0), the y-component of the magnetic field produced by these three wires at the origin?T
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3)
What is Fx(1), the x-component of the force exerted on a one meter length of the wire carrying current I1?N
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4)
What is Fy(1), the y-component of the force exerted on a one meter length of the wire carrying current I1?N
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5)
What is Fx(2), the x-component of the force exerted on a one meter length of the wire carrying current I2?N
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6)
Another wire is now added, also aligned with the z-axis at (x,y) = (0, -32 cm) as shown. This wire carries current I4 A. Which of the following statements is true?
If I4 is directed along the positive z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin.
If I4 is directed along the negative z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin.
If I4 is directed along the positive z-axis, then it is possible to make the x-component of the magnetic field equal to zero at the origin.
If I4 is directed along the negative z-axis, then it is possible to make the x-component of the magnetic field equal to zero at the origin.
d IExplanation / Answer
d = sqrt(18.5^2+32^2) = 37 cm = 0.37 m
1)
magnetic field due to I1 , B1x = 0
magnetic field due to I2 , B2x = 0
magnetic field due to I3 , B3x = +uo*I3/(2*pi*r3)
r3 = 32 cm = 0.32 m
magnetic field due to I3 , B3x = 4*pi*10^-7*6.9/(2*pi*0.32) = 4.31*10^-6 T
Bx = B1x + B2x + B3x = 4.31*10^-6 T
(2)
magnetic field due to I1 , B1y = -uo*I1/(2*pi*d/2) = -4*pi*10^-7*2/(2*pi*0.185) = -2.16*10^-6 T
magnetic field due to I2 , B2y = -uo*I2/(2*pi*d/2) = -4*pi*10^-7*1.3/(2*pi*0.185) = -1.41*10^-6 T
magnetic field due to I3 , B3y = 0
By = B1y + B2y + B3y = -3.57*10^-6 T
====================================
3)
x component
force due to I2 , F2x = -uo*I2*I1*l/(2*pi*d) = -4*pi*10^-7*1.3*2*1/(2*pi*0.37) = -1.41 *10^-6 N
force due to I3, F3x = -uo*I3*I1*cos60*l/(2*pi*d) = -4*pi*10^-7*6.9*2*cos60*1/(2*pi*0.37) = -3.73*10^-6 N
F1x = F2x + F3x = -5.14*10^-6 N
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4)
y component
force due to I2 , F2y = 0
force due to I3, F3y = -uo*I3*I1*sin60*l/(2*pi*d) = -4*pi*10^-7*6.9*2*sin60*1/(2*pi*0.37) = -6.46*10^-6 N
F1y = -6.46*10^-6 N
=========================================
5)
x component
force due to I1 , F1x = uo*I2*I1*l/(2*pi*d) = 4*pi*10^-7*1.3*2*1/(2*pi*0.37) = 1.41 *10^-6 N
force due to I3, F3x = -uo*I3*I2*cos60*l/(2*pi*d) = -4*pi*10^-7*6.9*1.3*cos60*1/(2*pi*0.37) = -2.42*10^-6 N
F2x = F1x + F3x = -1.01*10^-6 N
=============================
6)
If I4 is directed along the positive z-axis, then it is possible to make the
x-component of the magnetic field equal to zero at the origin.
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