Question

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I₂ = 11.5 A. Point A is the midpoint between the wires. The total distance between the wires is d = 10.2 cm. Point C is 5.75 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 2.11 m long segments of the wires?

Answer 1

A)

Distance of the point from the 1st wire carrying current I1 = r1 = 10.2 + 5.75 cm

r1 = 15.95 cm = 0.1595 m ;

Distance of the point from the 2nd wire carrying current I2 = r2 = 5.75 cm = 0.0575 m

I2 = 11.5 A

Net Magnetic Induction at the point due to both the wires = 2K{I1/r1 - I2/r2} = 0,

where K = μo/4π Wb/A-m = 10^(- 7) Wb/A-m

I1 = I2*(r1/r2) = 11.5*( 0.1595/0.0575) A

I1 = 31.9 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 = 10.2/2 cm = 5.1 cm = 0.051 m

Resultant Magnetic Induction at A = B = B1 + B2 = (2K/r)*(I1 + I2)

= {10^(- 7)}*2(11.5 + 31.9) / 0.051

= {10^(- 7)}*2*(43.4)/0.051

= 1.7 x 10^- 4 Wb/m^2

C)

Mutual Force per unit length of the wire = F/L = 2K(I1)(I2)/d,

where d = 10.2 cm = 0.102 m

Length of the wire segments = L = 2.11 m

F = 2KL(I1)(I2)/d

F = 2*{10^(- 7)}*(2.11)*(31.9)*(11.5) / (0.102)N

F = 1.52 x 10^-2 N