As shown in the figure below, two long parallel wires (1 and 2) carry currents o
ID: 2038004 • Letter: A
Question
As shown in the figure below, two long parallel wires (1 and 2) carry currents of I1- 3.22 A and I2 - 5.15 A in the direction indicated. (a) Determine the magnitude and direction of the magnetic field at a point midway between the wires (d 10.0 cm) 772 What is the ri for the magnitude of the magnetic field due to a long, straight wire? Knowing each magnetic field vector, how must the vectors be added to fid the net field? HT 90 ght-hand rule for currents? What is the direction of the magnetic field vector due to current 11? Due to current 12? What is the formula magnitude direction counterclockwise from the +x-axis (b) Determine the magnitude and direction of the magnetic field at point P, located d 10.0 cm above wire 1. 728.4 What is the for the magnitude of the magnetic field due to a long, straight wire? Knowing each magnetic field vector, how must the vectors be added to find the net field? ?? right-hand rule for currents? What is the direction of the magnetic field vector due to current 11? Due to current 12? What is the formula magnitude direction What are the x and y components of the net magnetic field? Can you use trigonometry to determine the direction?0 counterclockwise from the +x-axisExplanation / Answer
Field of I1 is B1= mu0*I1 / 2*pi*d/2
where mu0 = 1.25*10^-6,
So, B1 = 1.288*10^-5 T clockwise
Similarly, B2 = mu0*I2 / 2*pi*d/2 = 2.06*10^-5 T counterclockwise
So, field at a point midway between the wires = B2- B1 = 7.72*10^-6 T counterclockwise
Now, B1 field at P due to I1 = mu0*I1/ 2*pi*1.41d = 4.5*10^-6 along tan^-1(0.1/0.1) = 45 degrees counterclockwise
Similarly, B2 = mu0*I2/ 2*pi*d = 1.03*10^-5 T along x axis
So, net field is (B1^2+B2^2+2B1B2cos45)^1/2 = 1.3852*10^-5 T along tan^-1(B1sin45/B2+B1 cos45) = 13.28 degrees counterclockwise
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