Two straight parallel wires carry currents in opposite directions as shown in th
ID: 1873353 • 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 = 10.2 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?
Explanation / Answer
Solution:-
A) 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 = 5.03 cm = 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 = o/4 Wb/A-m = 10^(- 7) Wb/A-m
I1 = I2*(r1/r2) = 10.2*( 0.1713/ 0.0503 ) A = 34.74 A
B) At the point A :
The two magnetic fields are in the same direction
Magnetic Induction from 1st wire = B1 = 2K(I1/r)
Magnetic Induction from 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 + 34.74)/0.0605
= 1.485*10^(- 4) Wb/m²
C) 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 = 1.11 m
F = 2KL(I1)(I2)/d
= 2*{10^(- 7)}*(1.31)*(34.74)*(10.2)/(0.121) N = 7.67*10^(- 4) N
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