Question

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I₂ = 10.2 A. Point A is the midpoint between the wires. The total distance between the wires is d = 12.1 cm. Point C is 5.03 cm to the right of the wire carrying current I₂. Current I₁ is adjusted so that the magnetic field at C is zero. Calculate the value of the current I₁.Calculate the magnitude of the magnetic field at point A.What is the force between two 1.31 m long segments of the wires?

Answer 1

For the current I1

Distance of the point from the 1st wire carrying current I1 = r1

= 12.1 + 5.03 cm
= 17.13 cm

= 0.1713 m ;
Distance of the point from the 2nd wire carrying current I2 = r2

= 5.03cm

= 0.0503 m
I2 = 10.2 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)

= 10.2*( 0.1713/0.0503) A

= 34.736 A

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 = 12.1/2 cm

= 6.05 cm

= 0.0605 m
Resultant Magnetic Induction at A = B

= B1 + B2

= (2K/r)*(I1 + I2)
= {10 ^-7}*2(10.2 + 34.736)/0.0605
= {10 ^-7}*2*(44.936)/0.0605
= 1.485*10^{- 4} Wb/m²

Mutual Force per unit length of the wire = F/L

= 2K(I1)(I2)/d,

where d = 12.1 cm = 0.121 m
Length of the wire segments = L = 1.31 m
F = 2KL(I1)(I2)/d
= 2*{10 ^-7}*(1.31)*(34.736)*(10.2)/(0.121) N

= 7.67*10^{- 4} N