Question

Three parallel wires that are each 500 m long and spaced 0.6 m apart, carry currents and exert forces upon each other. Wire #1 and wire #2 are both fixed in position and cannot move. Wire #3 can move, but will break if it experiences a force of 20 N or more. Wire #1 is carrying a current of 120 A to the left and wire #2 is carrying a current of 240 A to the right.

A) What is the magnitude of the current that wire #3 must carry in order to experience enough force to break it?

B) 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 enough force to break it?

C) Suppose the current in wire #1 is suddenly shut off but the current in wire #2 remains the same. What is the magnitude of the current that wire #3 must carry in order to experience enough force to break it?

Answer 1

$B= \mu_o I/2 \pi r$

magnetic field due to first wire in position of third wire

I=120 , r=1.2m

B=0.02 x 10-3 tesala

magnetic field due to second wire

I=240 , r =.6m

B2=-0.08 x 10-3 tesala

Net magnetic field

B=B1-B2=0.06 x 10^-3 tesala

force in wire

F=iLB

I=666.667 ampere

B) if current is half

B1=.00001

Bnet=0.00007

and I=571.4

C) B1=0

Bnet =0.00008

I=500 A