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Magnetic Field Of Two Parallel Wires

  • Thread starter rstlr
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  • #1
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A long wire carrying a 5.0A current perpendicular to the xy-plane intersects the x-axis at x=-2.0 cm. A second, parallel wire carrying a 3.0A current intersects the x-axis at x=2.0 cm. At what point or points on the x-axis is the magnetic field zero, (a) if the currents are in the same direction (b) if the currents are in opposite directions.


B = (u * I)/(2Pi * d)?

Im not really sure how to do this :(
 

Answers and Replies

  • #2
berkeman
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A long wire carrying a 5.0A current perpendicular to the xy-plane intersects the x-axis at x=-2.0 cm. A second, parallel wire carrying a 3.0A current intersects the x-axis at x=2.0 cm. At what point or points on the x-axis is the magnetic field zero, (a) if the currents are in the same direction (b) if the currents are in opposite directions.


B = (u * I)/(2Pi * d)?

Im not really sure how to do this :(
Welcome to the PF.

Start with Ampere's Law, and use the Right Hand Rule to figure out the directions of the two B-fields from the t wires at poins along the x axis... Please show us your work....
 
  • #3
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Thanks!

Ok... so basically I have to guess at numbers till I find it?... that doesn't seem like a fun time :/
 
  • #4
berkeman
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56,656
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Thanks!

Ok... so basically I have to guess at numbers till I find it?... that doesn't seem like a fun time :/
Nope. No guessing involved. You sort of wrote out Ampere's Law in your original post (OP). The only thing missing is the Right Hand Rule (which is kind of hard to write on 2-D paper). You should be able to draw a sketch of the B-field for each wire (how does the B-field circulate around each wire?), and think some about how the two B-fields add. Can you post a scan of the drawing for the homework problem? Or else at least your sketch.

Don't guess, learn how to think about these problems in 3-D. That's an important skill in this course that you're taking now, and in many of your future courses.
 
  • #5
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Ok... Well based on what I had drawn out earlier, if the currents were in the same direction, it would be 0 = B1 - B2 in the center. Where B1 is the 5A current and B2 is the 3A current.

Then I assume I would solve for distance (d) so it would look something like

0 = (4pi x 10^-7)(5A)/(2pi)(d) - (4pi x 10^-7)(3A)/(2pi)(d)

Since they have common denominators I would just multiply each side by d? getting

d = (4pi x 10^-7)(5A)/(2pi) - (4pi x 10^-7)(3A)/(2pi)
d = 4x10^-7 m

which obviously isn't right, so I must be doing something seriously wrong. (the answer for when they are in parallel is supposed to be .5 cm)
 
  • #6
berkeman
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Ok... Well based on what I had drawn out earlier, if the currents were in the same direction, it would be 0 = B1 - B2 in the center. Where B1 is the 5A current and B2 is the 3A current.

Then I assume I would solve for distance (d) so it would look something like

0 = (4pi x 10^-7)(5A)/(2pi)(d) - (4pi x 10^-7)(3A)/(2pi)(d)

Since they have common denominators I would just multiply each side by d? getting

d = (4pi x 10^-7)(5A)/(2pi) - (4pi x 10^-7)(3A)/(2pi)
d = 4x10^-7 m

which obviously isn't right, so I must be doing something seriously wrong. (the answer for when they are in parallel is supposed to be .5 cm)
They don't have common denominators. The two distances to the 0 point from each wire will be different if the currents are different... Otherwise you are on the right track.
 
  • #7
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are we solving for d? OR do you use the distances from each individual wire to point 0 as d?
 

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