Two long, parallel wires carry currents of I1 = 3.12 A and I2 = 4.65 A in the di
ID: 1472848 • Letter: T
Question
Two long, parallel wires carry currents of I1 = 3.12 A and I2 = 4.65 A in the direction indicated in the figure below. (Choose the line running from wire 1 to wire 2 as the positive x-direction and the line running upward as the positive y-direction.)
(a) Find the magnitude and direction of the magnetic field at a point midway between the wires (d = 20.0 cm).
(b) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 4.65-A current.
Explanation / Answer
a) at mid pioint, Bnet = B2 - B1
= mue*I2/(2*pi*d/2) - mue*I1/(2*pi*d/2)
= 4*pi*10^-7*4.65/(2*pi*0.2/2) - 4*pi*10^-7*3.12/(2*pi*0.2/2)
= 3.06*10^-6 T
direction : 270 degrees
b)
B1 = mue*I1/(2*pi*sqrt(2)*d)
= 4*pi*10^-7*3.12/(2*pi*sqrt(2)*0.2)
= 2.206*10^-6 T
B1x = -B1*cos(45) = -1.56*10^-6 T
B1y = B1*sin(45) = 1.56*10^6 T
B2 = mue*I2/(2*pi*d)
= 4*pi*10^-7*4.65/(2*pi*0.2)
= 4.65*10^-5 T
B2x = -4.65*10^-5 T
B2y = 0
Bnetx = B1x + B2x
= -6.21*10^-6 T
Bnety = B1y + B2y
= 1.56*10^-6 T
Bnet = sqrt(Bnetx^2 + Bnety^2)
= sqrt(6.21^2 + 1.56^2)*10^-6
= 6.4*10^-6 T
direction : theta = 180 - tan^-1(Bnety/Bnetx)
= 180 - tan^-1(1.56/6.21)
= 165.9 degrees
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