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

a) NO,   YES that cable 3 should be placed to the left of cable 1

b)  a = 8cm   the current is UP

Explanation:

The force between two cables that carries a current is given through the magnetic field created between them, it has the expression

        F₁ = μ₀ I₁ I₂ / 2π a l

where I₁ and I₂ are the current in each wire, at the separation between them and l the length of wire 1 on which the force is applied.

The direction of the force this given is given by the vector product between the current in wire 1 and the field created by 2, but we can summarize it:

* if the two currents are in the same direction the force is attractive

* if the two currents are contrary, the force is repulsive

With these relationships we can examine the different situations presented

a) Between cable 1 and 3 the force is zero. If cable 1 the current is upward in cable 3 it must be downward, so that the force is repulsive

Between wire 2 and 3. If wire 2 the current is downward, in wire 3 the current must be upward, for a repulsive force.

We see that in the two conditions the current in cable 3 has different directions, which is impossible, therefore this situation cannot occur

the answer is NO

The only way this is possible is that cable 3 should be placed to the left of cable 1

Answer YES

b) and c) Your question is unclear, I interpret that you want the position of cable 3 so that the force is zero on cable 3

we will assume that the current in cables 1 and 2 is ascending and that in cable 3 is descending

We write Newton's second law at the point of cable 3, the acceleration is zero

                    F13 - F23 = 0

                    F13 = F23

Let's calculate the forces

                  F₁₃ = μ₀ I₁ I₃ /2π a    l

                  F₂₃ = μ₀ I₂ I₃ /2π (d + a)    l

we substitute and simplify

                 I₁ / a = I₂ / (d + a)

where we have assumed that the length of all cables is the same

We solve to find

                I₁ (d + a) = I₂ a

                a (I₂ - I₁) = I₁ d₁

                a = I₁ / (I₂ - I₁) d

let's calculate

                a = 1.20 / (4.2 -1.2) 20

                a = 8cm

to the left of wire 1 and the direction is the current is UP

so that it is an attractive force between 1 and 3 and a repulsive force between 2 and 3

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.