Question

Two long wires hang vertically. Wire 1 carries an upward current of 1.80 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.00 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?
Is it possible in more than one way?
(b) Describe the position of wire 3.
(c) Describe the magnitude and direction of the current in wire 3.

Answer 1

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer 2

(a) Yes, this situation is possible. No, It is not possible in more than one way.

(b) The position of the wire is at a distance of 16.3 cm, in the direction left of the wire

(c) The magnitude of the current in wire 3 (I3) is 3.267 A and in a direction pointing in the downward direction.

(a)

The diagrammatic illustration of the information provided in the question can be seen in the image attached below.

From the image: suppose the forces on the wire as a result of the two other wires are equal and opposite, then the third wire will experience no net force. Hence, we can say Yes,  the situation is possible.

Given that:

• I₁ = 1.80 A
• I₂ = 4.00 A

It will not be possible to have more than one way, this is because the force acting on the other two wires will have opposite directions but they would not have the same equal magnitude.

The force per unit length existing in-between two current-carrying parallel wires can be determined by using the formula:

$\mathbf{\dfrac{F}{l} = \dfrac{\mu_o I_1I_2}{2 \pi r} }$

where:

• $\mathbf{I_1,I_2 }$ = current passing via conductors
• r = distance in-between the parallel conductors

SInce F1 = F2, Then:

$\mathbf{\dfrac{\mu_o\times I_1\times I_3\times l}{2 \pi r}= \dfrac{\mu_o\times I_2\times I_3\times l}{2 \pi (r+0.2)} }$

$\mathbf{\dfrac{I_1}{r}= \dfrac{ I_2}{ (r+0.2)} }$

$\mathbf{\dfrac{1.80}{r_1}= \dfrac{ 4.00}{ (r_1+0.2)} }$

1.80(r₁ + 0.2) = 4.00r₁

1.80r₁ +0.36 = 4.00 r₁

0.36 = 2.2r₁

r₁ = 0.36/2.2

r₁ = 16.3 cm in the direction left of the wire.

Again, we are to determine the magnitude of the current and its direction in wire 3 (I₃).

From the image, the forces of F21 = F23

∴

$\mathbf{\dfrac{\mu_o \times I_1 \times I_2 \times l }{2 \pi r} = \dfrac{\mu_o \times I_2 \times I_3 \times l }{2 \pi r}}$

$\mathbf{\dfrac{\mu_o \times 1.8 \times 4.0 \times l }{2 \pi \times 0.2} = \dfrac{\mu_o \times 4 \times I_3 \times l }{2 \pi \times 0.363}}$

Making I₃ the subject by equating both equations together, we have:

$\mathbf{I_3 = \dfrac{1.8 \times 0.363}{0.2}}$

I₃ = 3.267 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

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