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What is the force between two perpendicular wire carrying current, one to the north and one to the east?

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closed as too localized by Waffle's Crazy Peanut, Manishearth Apr 16 '13 at 6:12

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What were you thinking? –  user12345 Mar 14 '13 at 9:38
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duplicate: physics.stackexchange.com/questions/56713/… –  Michael Brown Mar 14 '13 at 12:50
    
Please see our homework policy. We expect homework type problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) –  Manishearth Apr 16 '13 at 6:12
    
@MichaelBrown: Next time, vote to close as a duplicate (use the close menu) if you think it is one :) –  Manishearth Apr 16 '13 at 6:12

2 Answers 2

Hints:

  1. The magnetic field due to a current-carrying wire circulates around the wire, in other words, the magnetic field vectors point tangent to circles whose planes are perpendicular to the wire and whose centers are intersected by the wire.

  2. The magnetic force experienced by a moving charge $q$ (like those in line currents) with a velocity $\mathbf v$ in a field $\mathbf B$ is given by $$ \mathbf F = q\mathbf v\times\mathbf B $$

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If the wires were parallel, the force per unit length of wire is given by $F =\frac{\mu_0I_1I_2}{2\pi r}$. In the case that both wires have the same magnitude of current, but in opposite directions, the equation becomes: $F =\frac{\mu_0I^2}{2\pi r}$ and the force causes the wires to repel.

If the wires are perpendicular however, the force between them is 0.

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