Two straight parallel wires carry currents in opposite directions as shown in th

Question

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I_2 = 11.2 A. Point A is the midpoint between the wires. The total distance between the wires is d = 11.8 cm. Point C is 5.00 cm to the right of the wire carrying current I_2. Current I_1 is adjusted so that the magnetic field at C is zero. Calculate the value of the current I_1.Calculate the magnitude of the magnetic field at point A.What is the force between two 1.41 m long segments of the wires

Explanation / Answer

magnetic field due to I1 is B1 = mu_o*i1/(2*pi*r) = (2*10^-7*i1)/0.168

magnetic field due to I2 is B2 = mu_o*i2/(2*pir) = (2*10^-7*11.2)/(0.05)

B1 = B2

(2*10^-7*i1)/0.168 = (2*10^-7*11.2)/(0.05)

i1/0.168 = 11.2/0.05

i1 = 37.632 A

magneric field at A is B =B1+B2

B1 = mu_o*i1/(2*pi*r) = (2*10^-7*37.632)/(0.118/2)= 12.75*10^-5 T

B2 = mu_o*i1/(2*pi*r) = (2*10^-7*11.2)/(0.118/2) = 3.79*10^-5 T

then B at A is (3.79+12.75)*10^-5 = 1.654*10^-4 T

Force F = mu_o*i1*i2*l/(2*pi*r) = (2*10^-7*37.632*11.2*1.41)/(0.118) = 1.007*10^-3 N

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