Forces on current carrying wires

In summary: So it should be F21 = F1 l B21, where B21 is the magnetic field at location of wire 21. But it looks like it is not, because the field at wire 21 (indicated by the green dot) is not equal to the field at wire 1 (indicated by the blue dot). What am I doing wrong?
  • #1
flyingpig
2,579
1

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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  • #2
In what direction do you want to take x?
 
  • #3
To the right as in the picture
 
  • #4
then how will d+x be the position of green wire wrt blue wire ??
 
  • #6
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
 
  • #7
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 ?
 
  • #8
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
 
  • #9
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.
 
  • #10
No that one is answered, I made it complicated
 
  • #11
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
 
  • #12
What question are you trying to answer?
 
  • #13
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FAQ: Forces on current carrying wires

What is the relationship between the magnetic field and the force on a current carrying wire?

The force on a current carrying wire is directly proportional to the strength of the magnetic field it is in. This means that the stronger the magnetic field, the greater the force on the wire will be.

How does the direction of the current affect the force on a wire in a magnetic field?

The direction of the current will determine the direction of the force on the wire. If the current is flowing in the same direction as the magnetic field, the force will be in one direction. If the current is flowing in the opposite direction, the force will be in the opposite direction.

What is the right hand rule for determining the direction of the force on a current carrying wire?

The right hand rule states that if you point your thumb in the direction of the current and your fingers in the direction of the magnetic field, your palm will face in the direction of the force on the wire.

How does the length of a wire affect the force on it in a magnetic field?

The length of the wire does not affect the force on it in a magnetic field. The force is determined by the strength of the magnetic field and the current flowing through the wire, not the length of the wire itself.

Can the force on a current carrying wire be greater than the weight of the wire?

Yes, the force on a current carrying wire can be greater than its weight. This is because the force is determined by the strength of the magnetic field and the current, not the weight of the wire.

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