A wire, whose linear mass density is 150g/m, carries a current of 40A (supplied by a flexible negligible weight). This wire lies parallel to, and on top of, another horizontal wire on a table. What current must flow through the bottom wire in order to repel and support the top wire at a height of 4.0 cm above it? The top wire is held in place by frictionless guide plates.

Thanks,

...Dan

The field due to a wire is

$$B=\frac{\mu_{0}I}{2\pi r}$$

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

$$F_{mag}=qvB$$

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

$$F_{mag}=\frac{qLB}{\Delta t}$$

we can then move that $$\Delta t$$ over to $$q$$ and get somthing that looks like this

$$F_{mag}=ILB=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L$$

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

$$0=F_{mag}-F_{grav}$$

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

$$\mu gL=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L$$

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

$$I_{2}=\frac{\mu 2\pi rg}{\mu_{0}I_{1}}$$

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.

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Thanks for the help