Two straight parallel wires carry currents in opposite directions as shown in th
ID: 1536191 • Letter: T
Question
Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I2 = 10.3 A. Point A is the midpoint between the wires. The total distance between the wires is d = 11.8 cm. Point C is 4.51 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 1.41 m long segments of the wires?
Please explain
Explanation / Answer
A)
Distance of the point from the 1st wire carrying current I1 = r1 = 11.8 + 4.51 cm
= 1631 cm = 0.1631 m ;
Distance of the point from the 2nd wire carrying current I2 = r2 = 4.51 cm = 0.0451 m
I2 = 10.3 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) = 10.3x( 0.1631/0.0451) A = 37.249 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 = 11.8/2 cm = 5.90 cm = 0.0590 m
Resultant Magnetic Induction at A = B = B1 + B2 = (2K/r)*(I1 + I2)
= {10^(- 7)}*2(10.3 + 37.249)/0.0590
= 1.6118x10^(- 4) Wb/m²
C)
Mutual Force per unit length of the wire = F/L = 2K(I1)(I2)/d, where d = 11.8 cm = 0.118 m
Length of the wire segments = L = 1.41 m
F = 2KL(I1)(I2)/d
= 2x{10^(- 7)}x(1.41)x(37.249)x(10.3)/(0.118) N = 9.16x10^(- 4) N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.