Alright PF logged me out and erased all my work

[flyingpig]

Homework Statement

Alright I spent like 20 mins typing out the problem and PF logged me out and I couldn't hit backspace to save work when I hit submit thread so I apologize if this seemed a lack of effort...

Problem

Two long, parallel wires are attracted to each other by a force per unit length of 320 μN/m when they are separated by a vertical distance of 0.500m. The current in the upper wire is 20.0 A to the right. Determine the location of the line in the place of the two wires along which the total magnetic field is zero.


http://img140.imageshack.us/img140/7098/bfield.th.png

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The Attempt at a Solution

So I did

|F_{12}| = |-F_{21}|

\frac{F_{12}}{l} = \frac{F_{21}}{l}

I_2 B_1 = I_1 B_2

I_2 \frac{\mu_0 I_1}{2\pi (0.5 - y)} = I_1 \frac{\mu_0 I_2}{2\pi y}

Solving I get

0.5 - y = y

y = 0.25 which is wrong. The actual answer is 0.333m

 

[SammyS]

Infuriating !!! Isn't it?

 

[Caramon]

Give me a minute, I'll solve this.

EDIT: Before I do anything. Are you sure that your expression here is correct?:
<br /> I_2 B_1 = I_1 B_2<br />

When I expand from:
<br /> |F_{12}| = |-F_{21}|<br />

I get:
<br /> I_1 l_1 B_1 = I_2 l_2 B_2<br />
Where the lengths of the wires can cancel since they are the same variable.

So the resulting equation where you can sub your mu-sub-zero/2pi, etc. would be:
<br /> I_1 B_1 = I_2 B_2<br />

Let me know if this makes sense.

 

[SammyS]

What is meant by F₁₂ & F₂₁.

Be more explicit than usual.

 

[flyingpig]

The force exerted by 1 onto 2 is supposed to be F₁₂, same with F₂₁

 

[SammyS]

The force exerted by 1 onto 2 is supposed to be F₁₂, same with F₂₁

Of course we know that these forces are equal (in magnitude), and opposite. ---Issac Newton tells us that, even though he knew nothing of these equations relating current and B field.

What the problem wants from you is to:

Determine the location of the line along which the total magnetic field due to the two wires is zero.​

Yes, I did change the wording so the problem makes sense. If the problem is different, in a significant way, than what I said, please change it to the wording of your textbook.

 

[flyingpig]

I know that's why I set the force per unit length equal

 

[SammyS]

There are two main steps needed to solve this.

First, you need to find the current in the lower wire. This is similar to a problem you've already done. Here you know the force per unit length & its direction, the distance between wires and the current in one of the wires.

Previously, you were given both currents & the separation -- and were asked for the force per unit length.

***********

The second part is to find the location where the B field due to the two wires cancels (is zero).

 

[flyingpig]

OKay so I₁ is

\frac{F_{12} 2\pi y}{l\mu_0 I_2} = I_1

So numerically it is

 

[flyingpig]

Please...need more help

 

[SammyS]

\frac{F_{12}}{\ell}=320\,{\mu}\text{N/m}

y=0.500\,\text{m}

I_2=20.0\,\text{A}

Calculate I₁.

 

[flyingpig]

I₁ = 40a

 

[SammyS]

In what direction is I₁, if the wires are attracted to each other?

 

[flyingpig]

Same direction as I₂ as I have drawn in the picture.

 

[SammyS]

Yes, you did draw it in your picture. It's helpful to make your answer as complete as is reasonably possible.

Now, for the main question: (Remember, here you are only concerned with B field, NOT force.)
Notice that the magnetic field due to wire 1 is out of the page (positive, in my view) anywhere above wire 1 (y > 0.500 m) & the magnetic field due to wire 1 is into the page (negative, in my view) anywhere below wire 1 (y < 0.500 m) .

Similarly, the magnetic due to wire 2 is positive for y > 0, & negative for y < 0.

So, the only place where adding them together can cancel them, is: ______ ?

 

[flyingpig]

y ∈ (0,0.5)

 

[SammyS]

OK, solve the problem.

 

[flyingpig]

Yup I got y = 1/3. I was close though...I mean I had the idea of setting something equal to something.

Thanks again Sammy lol.