11-10 points SerCP11 19.P050. My Notes Ask Your Teacher Two long, parallel wires
ID: 1577681 • Letter: 1
Question
11-10 points SerCP11 19.P050. My Notes Ask Your Teacher Two long, parallel wires carry currents of I1-3.14 A and 12-4.80 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) magnitude direction from the positive x-axis (b) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 4.80-A current. magnitude direction to the left of the vertical Need Help? Read ItExplanation / Answer
(A) due to I1, B1 = u0 I1 / (2 pi d/2) upward
due to I2. B2 = u0 I2 / ( pi d) downward
B = B2 - B1 = u0 (I2 - I1) / (pi d )
= (4pi x 10^-7)(4.80 - 3.14) / ( pi 0.20)
= 3.32 x 10^-6 T downward
magnitude = 3.32 x 10^-6 T Or 3.32 uT
direction = 90 deg
(B) due to I2, B2 = (u0 I2 / 2 pi d )(-i)
= - 4.8 x 10^-6 i
due to I1, B1 = (u0 I1 / 2 pi d)( -cos45i + sin45j)
B1 = - 2.22 x 10^-6 i + 2.22 x 10^-6 j
B = B1 + B2 = - 7.02 x 10^-6 i + 2.22 x 10^-6 j
magnitude = sqrt(7.02^2 + 2.22^2) x 10^-6
= 7.36 x 10^-6 T Or 7.36 uT
direction = tan^-1(7.02/2.22) = 72.45 deg
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