Two straight parallel wires carry currents in opposite directions as shown in th
ID: 1476368 • 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 = 11.0 A. Point A is the midpoint between the wires. The total distance between the wires is d = 10.2 cm. Point C is 5.14 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 2.57 m long segments of the wires?
Explanation / Answer
A) Distance of the point from the 1st wire carrying current I1 = r1 = 10.2 + 5.14 cm
= 15.34 cm = 0.1534 m ;
Distance of the point from the 2nd wire carrying current I2 = r2 = 5.14 cm = 0.0514 m
I2 = 11 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) = 11*( 0.1534/0.0514) A = 32.83 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 + 32.83)/0.051 = 17.1882*10-5 Wb/m²
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 = 2.57 m
F = 2KL(I1)(I2)/d
= 2*{10-7}*(2.57)*(32.83)*(11)/(0.102) N = 1.8198*10^(- 3) N
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