Two long, parallel wires carry currents of I 1 = 2.96 A and I 2 = 5.25 A in the
ID: 1422441 • Letter: T
Question
Two long, parallel wires carry currents of I1 = 2.96 A and I2 = 5.25 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 5.25-A current.
° to the left of the vertical
**WILL RATE! PLEASE HELP**
magnitude 4.52 T direction -90° from the positive x-axisExplanation / Answer
(a)
B = uo/(2*pi*d/2) * [I2 - I1]
B = (4*10^-7)/ (0.2) * [5.25 - 2.96] T
B = 4.58 uT
(b)
B2 = uo *I2 /(2*pi*d) i^
B2 = (4*10^-7*5.25)/(2*0.2) j^
B2 = 5.25 uT
Distance btw I1 & P, = sqrt(d^2+d^2) = d*sqrt(2)
B1 = uo *I1 /(2*pi*d*sqrt(2)) * cos(45) i^ + uo *I1 /(2*pi*d*sqrt(2)) * sin(45) j^ T
B1 = (4*10^-7*2.96)/(2*0.1*sqrt(2)) * cos(45) i^ + (4*10^-7*5.40)/(2*0.2*sqrt(2)) * sin(45) j^
B2 = 1.48 uT i^ + 1.48 uT j^
Net Magnetic Field,
B = sqrt((5.25 + 1.48)^2 + 1.48^2) uT
B = 6.89 uT
Direction, = tan^-1(6.73/1.48)
Direction, = 77.6 o to the left of the vertical.
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