## Forces on current carrying wires

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.

(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

$$\left |\vec{B_1} | \right= \frac{\mu_{0} I_{1}}{2\pi d}$$

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?

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

Look at picture

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

$$\left | \vec{F_{12}} \right | = I_{1} l \left| \vec{B_1} \right|$$

$$\left | \vec{F_{12}} \right | = I_{1} \frac{\mu_{0} I_{1}}{2\pi d}$$

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

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

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 Mentor In what direction do you want to take x?
 To the right as in the picture

## Forces on current carrying wires

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

 Not sure what you mean, but I added a picture http://img805.imageshack.us/i/asasss.png/
 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 = μoi2/2πd

 Quote by flyingpig Not sure what you mean, but I added a picture http://img805.imageshack.us/i/asasss.png/
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 ???

 Quote by 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 = μoi2/2πd
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.

 Quote by 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 ???

 Mentor 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.

 Quote by 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
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

 Mentor What question are you trying to answer?
 Uploaded with ImageShack.us 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