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webassign.net JavaScript Dates 9) Ohmkar Hlarious Comments On Vennala Kishor l Hw#11-Sources of magnetic field Two Long, Parallel Wires Carry Currents Of n = 7. 1/4 points | Previous Answers GianPSEWA4 28.P.003. My Notes As shown in the figure below, two long parallel wires (1 and 2) carry currents of 11-2.96 A and 12 = 5.35 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) magnitude 9.56 direction HT o 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 1.646 What is the right-hand rule for currents? What is the direction of the magnetic field vector due to current I1 Due to current I2? What is the formula 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? T magnitude direction o counterclockwise from the +x axis Submit Answer Save Progress Practice Another VersionExplanation / Answer
(a)
I1 = 2.96 A
I2 = 5.35 A
r = d/2 = 5 cm = 0.05 m
magnetic field due I1 = B1y = -uo*I1/(2*pi*r) = -4*pi*10^-7*2.96/(2*pi*0.05) = -1.184*10^-5 T
magnetic field due I2 = B2y = uo*I2/(2*pi*r) = 4*pi*10^-7*5.35/(2*pi*0.05) = 2.14*10^-5 T
Bnet = B1y+ B2y = 0.956*10^-5 = 9.56 uT <<<<-------ANSWER
direction 90 <<<<-------ANSWER
===============
(b)
magnetic field due I1 = B1x = -uo*I1/(2*pi*r1)
r1 = d
B1x = 4*pi*10^-7*2.96/(2*pi*0.1) = 5.92*10^-6 T
B1y = 0
r2 = sqrt(d^2+d^2) = 14.14 cm = 0.1414 m
B2x = uo*I2*cos45/(2*pi*r2)
B2x = 4*pi*10^-7*5.35*cos45/(2*pi*0.1414) = 5.35*10^-6 T
B2y = uo*I2*sin45/(2*pi*r2)
B2y = 4*pi*10^-7*5.35*cos45/(2*pi*0.1414) = 5.35*10^-6 T
Bx = B1x + B2x = 11.27 uT
By = B1y + B2y = 5.35 uT
B = sqrt(By^2 + Bx^2) = 12.5 uT <<<<-------ANSWER
direction tan^-1(By/Bx)
direction = 25.4 degrees <<<<<-------ANSWER
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