Two long, parallel wires carry currents of I 1 = 3.28 A and I 2 = 4.55 A in the
ID: 1429812 • Letter: T
Question
Two long, parallel wires carry currents of I1 = 3.28 A and I2 = 4.55 A in the direction indicated in the figure below. (Choose the line running from wire 1 to wire 2 as the positive x-axis and the line running upward from wire 1 as the positive y-axis.)
(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 4.55-A current.
Explanation / Answer
A. At midpoint
B1 = u0*I/2*pi*r = 2*10^-7*3.28/0.1 = 6.56*10^-6 T (upward)
B2 = u0*I/2*pi*r = 2*10^-7*4.55/0.1 = 9.1*10^-6 T (downward)
Net B = (9.1 - 6.56)*10^-6 = 2.54*10^-6 T (downward)
(B). At point P above wire 2
wire 1, wire 2 and point P will form a right-angle triangle whose three sides will be
AB = 20cm, BP = 20 cm and AP = 28.3 cm
B1 = u0*I/2*pi*r = 2*10^-7*3.28/0.283 = 2.31*10^-6 T
B2 = u0*I/2*pi*r = 2*10^-7*4.55/0.2 = 4.55*10^-6 T
Bx = (4.55 + 2.31*cos 45 deg)*10^-6 = 6.18*10^-6
By = 2.31*10^-6*sin 45 deg = 1.63*10^-6
B = sqrt(Bx^2 + By^2) = sqrt(6.18^2 + 1.63^2)*10^-6
B = 6.39*10^-6 T
direction = arctan(1.63/6.18) = 14.77 deg
Comment below if you have any doubt.
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