Magnetic field due to two wires

  • Thread starter Siune
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  • #1
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Homework Statement


We have wires like in the picture and we want to know the value of magnetic field as
function of x:

NOTE: We want to know the magnetic field at point P ( P is on +x axis ) ( Which is totally arbitary!) The picture is there for just to give the idea.

189576A.jpg


R is the distance from each wire to point P.

Homework Equations



[itex] H = \frac{I}{2 \pi R }[/itex]

The Attempt at a Solution



Okey, so the currents are going in opposite ways. By the right hand rule we get the direction of the magnetic field due to each wire!

We notice that the y-component cancels out ( due to symmetry ), and that the magnetic field is only the x component due to each wire at point P.

So by superposition principle we get the magnetic field at point P on x -axis:

[itex]H = sin(\theta) \frac{ I }{ \pi R }[/itex]

Now [itex]sin(\theta)[/itex] is the angle between x axis and the distance R from each wire.

So we get from trigonometry:

[itex]H = \frac{d \cdot I }{ \pi R^2}[/itex]

[itex]H = \frac{d \cdot I }{ (x ^2 + d ^2 ) \pi }[/itex]

Not sure if this is right? At least the units match ( A / m ), but... problem is:

My assigment gives us only the current I and the x -axis distance to point P? So I assume the right answer of the magnetic field at point P should be independent of d? But that doesn't make sense to me.

Sincerely yours,
Siune
 

Answers and Replies

  • #2
TSny
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We notice that the y-component cancels out ( due to symmetry ), and that the magnetic field is only the x component due to each wire at point P.

So by superposition principle we get the magnetic field at point P on x -axis:

[itex]H = sin(\theta) \frac{ I }{ \pi R }[/itex]

Now [itex]sin(\theta)[/itex] is the angle between x axis and the distance R from each wire.
If θ is the angle between the x-axis and R, make sure you are using the correct trig function for getting the x-component.
 
  • #3
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Hmm, as the distance vector R from one wire and magnetic field from same wire are perpendicular I get that the x component of magnetic field by that wire

so [itex]sin(\theta) = \frac{d}{R}[/itex]

and

[itex]R^2 = x^2 + d^2 [/itex]
?
 
  • #4
TSny
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Sorry, my mistake. You are entirely correct. The answer does depend on both x and d.
 
  • #5
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No problem at all, wouldn't have been the first time I make the mistake on the easiest part of the problem. ^^
 

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