Two long, parallel conductors carry currents I1 " 3.00 A and I2 " 3.00 A, both directed into the page in Figure P30.15. Determine the magnitude and direction of the resultant magnetic field at P?

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Answer 1 · 2017-12-01T23:29:18

The polar form is:

#vec(B_1)+ vec(B_2) = (30(mu_0)/(2pi), pi)#

We need the direction of #vec(R_1)#.

We observe that this is a 5,12,13 right triangle, therefore, the angle, #alpha_1#, between the vertical and #vec(R_1)# is:

#alpha_1 = cos^-1(5/13)#

The direction of #vec(R_1)# is:

#theta = cos^-1(5/13) - pi/2#

#theta ~~ -0.39479 #

The polar form of #vec(R_1)# is #(0.05" m", -0.39479)#

Using #vec(B_1) = (mu_0vec(I_1))/(2pivec(R_1)#

#vec(B_1) = (mu_0(-3" A", 0))/(2pi(0.05" m",-0.39479))#

Convert to rectangular form:

#vec(B_1) = (mu_0(-3" A"))/(2pi(0.05" m"))(cos(0.39479)hati+sin(0.39479)hatj)#

#vec(B_1) = (mu_0)/(2pi)(-55.4hati-23hatj)#

The direction of #vec(R_2)# is:

#theta = pi/2 - cos^-1(12/13)#

#theta ~~ 1.176#

The polar form of #vec(R_2)# is #(0.12" m", -0.39479)#

Using #vec(B_2) = (mu_0vec(I_2))/(2pivec(R_2)#

#vec(B_2) = (mu_0(-3" A", 0))/(2pi(0.12" m",1.176))#

#vec(B_2) = (mu_0(-3" A"))/(2pi(0.12" m"))(cos(-1.176)hati+sin(-1.176)hatj)#

#vec(B_2) = (mu_0)/(2pi)(25.4hati+23hatj)#

#vec(B_1)+ vec(B_2) = (mu_0)/(2pi)(-30hati)#

The polar form is:

#vec(B_1)+ vec(B_2) = (30(mu_0)/(2pi), pi)#