Forces on current carrying wires

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flyingpig
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Homework Statement



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.

http://img52.imageshack.us/img52/7425/asasssf.th.png

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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 on I2?





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?

http://img805.imageshack.us/img805/6568/asasss.th.png

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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

http://img854.imageshack.us/img854/6568/asasss.th.png

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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
 
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To the right as in the picture
 
then how will d+x be the position of green wire wrt blue wire ??
 
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
 
flyingpig said:
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 ?
 
cupid.callin said:
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.

cupid.callin said:
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 ?

Oh okay, that is answered
 
No that one is answered, I made it complicated
 
flyingpig said:
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