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 = 10.1 A. Point A is the midpoint between the wires. The total distance between the wires is d = 10.8 cm. Point C is 5.33 cm to the right of the wire carrying current I2. Current I1 is adjusted so that the magnetic field at C is zero.

1. Calculate the value of the current I1. 3.06×101

2. Calculate the magnitude of the magnetic field at point A.

3. What is the force between two 1.39 m long segments of the wires? 7.95×10-4 N

Please help with number 2!

Explanation / Answer

A) Distance of the point from the 1st wire carrying current I1 = r1 = 10.8 + 5.33 cm
= 16.13 cm = 0.1613 m ;
Distance of the point from the 2nd wire carrying current I2 = r2 = 5.33 cm = 0.0533 m
I2 = 10.1 A
Net Magnetic Induction at the point due to both the wires = 2K{I1/r1 - I2/r2} = 0,
where K = o/4 Wb/A-m = 10^(- 7) Wb/A-m
I1 = I2*(r1/r2) = 10.1*( 0.1613/0.0533) A = 30.565 A

B) At the point A :
The two magnetic fields are in the same direction
Magnetic Induction due to 1st wire = B1 = 2K(I1/r)
Magnetic Induction due to 2nd wire = B2 = 2K(I2/r)
where r = d/2 = 10.8/2 cm = 5.4 cm = 0.054 m
Resultant Magnetic Induction at A = B = B1 + B2 = (2K/r)*(I1 + I2)
= {10^(- 7)}*2(10.1 + 30.565)/0.054
= {10^(- 7)}*2*(40.665)/0.054
= 1.51*10^(- 4) Wb/m²

C) Mutual Force per unit length of the wire = F/L = 2K(I1)(I2)/d, where d = 10.8 cm = 0.108 m
Length of the wire segments = L = 1.19 m
F = 2KL(I1)(I2)/d
= 2*{10^(- 7)}*(1.19)*(30.565)*(10.1)/(0.108) N = 6.8*10^(- 4) N

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