Two long wires hang vertically. Wire 1 carries an upward current of 1.20 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.20 A. A third wire, wire 3, is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (a) Is this situation possible?
A. Yes
B. No
Is it possible in more than one way?
A. Yes
B. No
(b) Describe the position of wire 3.
distance cm
direction: left of wire 1
(c) Describe the magnitude and direction of the current in wire 3.
magnitude
direction
down

Explanation:

A, yes

b,

The force(per unit length) on wire 1 by 2 is to the left and

= μo*I1*I2/2πr

= 2*10^-7 * 1.2 * 4.2/0.20

= 5.04*10^-6N/m

From third law we have the force on wire2 by wire 1 is 5.04*10^-6N/m and to the right

So the magnitude of the force on each wire by wire 3 must be 4.2*10^-6N/m

Since the current in wire 2 is > wire 1 then wire 3 must be closer to wire 1 than 2

and wire 3 must be to the left of wire 1

Let x be the distance from wire 3 to wire 1 so x + 0.20 = distance from wire 3 to wire 2

Now μo*I1*I3/2πx = μo*I3*I2/2π(x + 0.20)

so solving for x we get

I1/x = I2/(x+0.20)

or x +0.20 = 4x

so x = 0.20/3 = 0.0667m

Now μo*I1*I3/2πx = 4.2*10^-6

so I3 = 4.2x10^-6 * 0.0667/(2.0x10^-7 * 1.2) = 1.167 A and its direction is down

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer