Problems 8, 9, and 10 relate to the following diagram. Two long parallel wires 1
ID: 1503583 • Letter: P
Question
Problems 8, 9, and 10 relate to the following diagram. Two long parallel wires 13 cm apart each carry a current of 20.4. each directed out of the page. Let l_1 be the top current, and h be the bottom current. Find the magnitude and the direction of the magnetic field from the current at point P. Find the magnitude and the direction of the magnetic field from the current at point P. Determine the total magnetic field at point P. Determine the total magnetic field at point P. What would the magnetic Held be if the currents were directed into the page instead of out of the page?Explanation / Answer
The field from each wire makes concentric circles about the source wire. The individual fields at point P are perpendicular to the lines from the wires to point P. We define B1 as the field from the closer point, and B2 as the field from the further point
B1 = uoI / 2pir1 = (4pi x 10^-7 x 20) / (2pi x 0.050 m) = 0.8 x 10^-4 T
B2 = uoI / 2pir2 = (4pi x 10^-7 x 20) / (2pi x 0.120 m) = 3.33 x 10^-5 T
Bnet = B1 + B2
12^2 + 5^2 = 13^2
Bnet = sqrt( B1 + B2)
Bnet = uoI/2pi*sqrt(1/r1^2 + 1/r2^2)
= uoI / 2pir1*sqrt(1 + r1^2/r2^2)
= [(4pi x 10^-7 x 20) / (2pi x 0.050)]*sqrt(1 + (5/12)^2
= 0.867 x 10^-4 T
To find direction of Bnet we can calculate x and y components:
theta = tan^-1(5/12) = 22.6 degree.
Bnet(x) = B1cos(90 - 22.6) - B2cos(22.6)
Bnet(x) = (0.3074 x 10^-4) - (0.3074 x 10^-4)
Bnet(x) = 0 T
Bnet(y) = B1sin(67.4) - B2sin(22.6)
Bnet(y) = (0.7385 x 10^-4) + (0.1280 x 10^-4)
Bnet(y) = 0.8665 x 10^-4 T
Bnet = 0.8665 x 10^-4 T at 90 degree.
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