Three parallel wires that are each 600m long and spaced .5 m apart, carry curren

Question

Three parallel wires that are each 600m long and spaced .5 m apart, carry currents and exert forces upon each other. Wire #1 and wire #2 are bothe fixed in position and cannot move. Wire #3 can move, but will break if it experiences a force of 15 N or more. Wire #1 is carring a current of 240 A to the left and Wire #2 is carring a current of 140 A to the right.

Suppose the current in Wire #1 is halved while the current in Wire #2 remains the same. What is the magnitude of the current that Wire #3 must carry in order to experience enouh force to break it?

Explanation / Answer

The net force on wire 3 is

Fnet = uo I1I3L / 2piR1 - uoI2I3L / 2PiR2

Putting the values

2pi(15) = uoI3L( I1/R1-I2/R2)

2(3.14)(15) = 4(3.14)(10-7)(600) ( 240 / 1 - 140/0.5) I3

I3 = 3125A

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