- #1

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Thanks,

...Dan

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- Thread starter Dan17
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- #1

- 4

- 0

Thanks,

...Dan

- #2

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The field due to a wire is

[tex]B=\frac{\mu_{0}I}{2\pi r}[/tex]

and the direction can be determined by use of the right hand rule. The force associated with these fields is given by the lorentz force law

[tex]F_{mag}=qvB[/tex]

when the field and the motion are perpendicular. Lets look at this another way... lets say that the lorentz force law read

[tex]F_{mag}=\frac{qLB}{\Delta t}[/tex]

we can then move that [tex]\Delta t[/tex] over to [tex]q[/tex] and get somthing that looks like this

[tex]F_{mag}=ILB=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L[/tex]

Utilizing Newton's Second law, in order for one wire to repel and support the other the forces must be equal.

[tex]0=F_{mag}-F_{grav}[/tex]

so we set the forces equal. ([tex]\mu[/tex] and [tex]\mu_{0}[/tex] are not related in any kind of way,[tex]\mu[/tex] is the linear mass density)

[tex]\mu gL=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L[/tex]

and we get a result that is only dependant on the second current

[tex]I_{2}=\frac{\mu 2\pi rg}{\mu_{0}I_{1}}[/tex]

Before you start putting numbers in make sure you understand what I have said here. And remember to give the current the proper polarity as indicated by the right hand rule.

[tex]B=\frac{\mu_{0}I}{2\pi r}[/tex]

and the direction can be determined by use of the right hand rule. The force associated with these fields is given by the lorentz force law

[tex]F_{mag}=qvB[/tex]

when the field and the motion are perpendicular. Lets look at this another way... lets say that the lorentz force law read

[tex]F_{mag}=\frac{qLB}{\Delta t}[/tex]

we can then move that [tex]\Delta t[/tex] over to [tex]q[/tex] and get somthing that looks like this

[tex]F_{mag}=ILB=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L[/tex]

Utilizing Newton's Second law, in order for one wire to repel and support the other the forces must be equal.

[tex]0=F_{mag}-F_{grav}[/tex]

so we set the forces equal. ([tex]\mu[/tex] and [tex]\mu_{0}[/tex] are not related in any kind of way,[tex]\mu[/tex] is the linear mass density)

[tex]\mu gL=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L[/tex]

and we get a result that is only dependant on the second current

[tex]I_{2}=\frac{\mu 2\pi rg}{\mu_{0}I_{1}}[/tex]

Before you start putting numbers in make sure you understand what I have said here. And remember to give the current the proper polarity as indicated by the right hand rule.

Last edited:

- #3

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Thanks for the help

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