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 I2 = 10.9 A. Point A is the midpoint between the wires. The total distance between the wires is d = 10.5 cm. Point C is 4.95 cm to the right of the wire carrying current I2. Current I1 is adjusted so that the magnetic field at C is zero.

a) Calculate the value of the current I1.

b) Calculate the magnitude of the magnetic field at point A.

c)What is the force between two 1.31 m long segments of the wires?

Explanation / Answer

   Given data
  
   I1,I2 are in opposite directions, and the magnetic field is given by

   B = mue 0 I/ 2*pi*r     r is distance from the wire

   I1= ?, I2 = 10.9 A, point A is at 10.5/2 = 5.25 cm from each wire,
        point C is at 4.95 cm from I2 wire and 15.45 cm from I1.

       B at C from I2 = 4*pi*10^-7*10.94 / (2*pi*0.0495)= 4.420*10^-5 T
  
       B at C from I1 = 4*pi*10^-7*I1 / (2*pi*0.1545)= 4.420*10^-5 T

       I1 = 34.14 A

At point A
  

   B1 = B to the right = 4*pi*10^-7*34.14 / (2*pi*0.0525) =13.0*10^-5 T
    B2 = B to the left = 4*pi*10^-7*10.9 / (2*pi*0.0525) = 4.152*10^-5 T

total magnetic field at A is sum of magnetic fields B = B1+B2= 17.15*10^-7 T

c) Magnetic force F/delta L = mue 0 *I1*I2/ 2*pi*r = (4*pi*10^-7*10.9*34.14) /( 2*pi*0.0525)*1.31

           F = 18.57*10^-4

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