Two straight parallel wires carry currents in opposite directions as shown in th
ID: 1259687 • 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 = 12.3 cm. Point C is 5.41 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 1.15 m long segments of the wires?
Explanation / Answer
A) Distance of the point from the 1st wire carrying current I1 = r1 = 12.3 + 5.41 cm
= 17.71 cm = 0.1771 m ;
Distance of the point from the 2nd wire carrying current I2 = r2 = 5.41 cm = 0.0541 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*( 0.1771/0.0541) A = 36.0 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 = 12.3/2 cm = 6.15 cm = 0.0615 m
Resultant Magnetic Induction at A = B = B1 + B2 = (2K/r)*(I1 + I2)
= {10^(- 7)}*2(11 + 36)/0.0615
= 1.528*10^(- 4) Wb/m
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