Forces on current carrying wires

flyingpig

1. The problem statement, all variables and given/known data

Two long parallel conductors separated by 10.0 cm carry currents in the same direction. The first wire carries a current I₁ = 5.00 A, and the second carries I₂ = 8.00 A.



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(a) What is the magnitude of the magnetic field created by I₁ and at the location of I₂
(b) What is the force per unit length exerted by I[sub] on I₂?





3. The attempt at a solution



For (a) it is

[tex]\left |\vec{B_1} | \right= \frac{\mu_{0} I_{1}}{2\pi d}[/tex]

Then it is just a matter of plugging in the numbers and it should come out as 10^-5T

But my question is, why is it that we use r = 10.00cm? Shouldn't it be d + x?



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Now for (b), I am just confused with Newton's third Law. I just don't understand I am wrong.

Look at picture



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By the right hand rule - cross product - the force F₁ should be to the right, that is the Force exerted by wire 1 (which I forgot to label as the one on the right) on wire 2.

If I were to do the Math I get

[tex]\left | \vec{F_{12}} \right | = I_{1} l \left| \vec{B_1} \right| [/tex]

[tex]\left | \vec{F_{12}} \right | = I_{1} \frac{\mu_{0} I_{1}}{2\pi d}[/tex]

[tex]\frac{\left | \vec{F_{12}} \right |}{l} = \frac{\mu_{0} I_{1}^2}{2\pi d}[/tex]

Which is wrong, but it should be [tex]\frac{\left | \vec{F_{12}} \right |}{l} = \frac{\mu_{0} I_{1} I_{2}}{2\pi d}[/tex] by Newton's third law

SammyS

In what direction do you want to take x?

flyingpig

To the right as in the picture

cupid.callin

then how will d+x be the position of green wire wrt blue wire ??

flyingpig

Not sure what you mean, but I added a picture

http://img805.imageshack.us/i/asasss.png/

cupid.callin

if F₁₂ means force on wire 1 due to wire 2

then F₁₂ = i₁ l B₁ where B₁ is the field at i₁ (Not due to i₁)

So B₁ = μ_oi₂/2πd

cupid.callin

I know you added a pic ...

What is the magnitude of the magnetic field created by I₁ and at the location of I₂

It asks for b at i₂ ... so you must use the distance of i₂ from i₁

how did you came up with d+x ???

flyingpig

Yes, but isn't that what I have in http://img854.imageshack.us/i/asasss.png/?

It feels right to me, but I know it is wrong.

Oh okay, that is answered

SammyS

Question (a):
" What is the magnitude of the magnetic field created by I₁ and at the location of I₂ ?"

The location of I₂ IS a distance d from I₁. No need for x.

Seems very clear to me.

flyingpig

No that one is answered, I made it complicated

cupid.callin

in eqn F = iBl

B is the field at location of wire ... i is current in wire and l is length of wire

B at wire 1 is due to wire 2(and not wire 1 itself) so for expression of B you use i₂ as i2 produce the current

SammyS

What question are you trying to answer?

flyingpig

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This picture, it should be F₂₁, but I cannot convince myself that it is. By the cross product I get a force F₁ to the right, which is the force of 1 exerted onto 2