A) Two straight parallel wires carry currents in opposite directions as shown in

Question

A) Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I2 = 11.4 A. Point A is the midpoint between the wires. The total distance between the wires is d = 10.6 cm. Point C is 5.46 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. THE ANSWER IS 33.53 A

I am confused by the next two questions though

B) Calculate the magnitude of the magnetic field at point A.

C) What is the force between two 1.85 m long segments of the wires?

(Show work with numbers in formula)

Explanation / Answer

asumption:

left wire carries current I1 , right wire carries current I2, point C is at a distance of 5.46 cm from right wire and hence 10.6+5.46=16.06 cm from the left wire.
as the two wires carry current in opposite direction,

their fields will also be directed in opposite direction

so net field=(mu*I2/(2*pi*0.0546))-(mu*I1/(2*pi*0.1606))

given net field at C is zero.

then (mu*I2/(2*pi*0.0546))=(mu*I1/(2*pi*0.1606))

==>I1=I2*0.1606/0.0546=11.4*0.1606/0.0546=33.5319 A

part B:

point A is midway between the two wires i.e. at a distance of 10.6/2=5.3 cm from each wire

as current in the wires is of opposite direction, in space between the two wires, field due to both wires will be added.

hence net field=(mu*I2/(2*pi*0.053))+(mu*I1/(2*pi*0.053))

=(mu*11.4/(2*pi*0.053))+(mu*33.5319/(2*pi*0.053))

=1.6955*10^(-4) T

part c:

force between the two wires=mu*I1*I2*length/(2*pi*distance)

=4*pi*10^(-7)*33.5319*11.4*1.85/(2*pi*0.106)=1.3343*10^(-3) N

as current are in opposite direction, force is repulsive in nature.

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