Finding the magnetic field due to a current

  • Thread starter jumbogala
  • Start date
  • #1
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Homework Statement


What is the magnetic field strength and direction at a point 4.0 cm above a wire that carries 12 A of current (the current goes toward the left).


Homework Equations


Magnetic field due to a current = {(1.26E-6 Tm/A)(I)(s x r)} / {(4pi)(r2)}

Note : s x r is a vector cross product and r here should have a hat over it.

The r in the denominator is just a radius.


The Attempt at a Solution


I am looking at the formula above and I'm not sure what s and r hat are.

Radius I guess would be 4.0 cm, and I think r hat would just be straight down (because doesn't it point from wherever you're calculating the field to the wire?)

But s... I have no idea what that is. Any ideas?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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s is the small element of the wire and r is the unit vector towards the point where the magnetic field is required. The direction of the magnetic field is perpendicular to s and r.
 
  • #3
423
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Okay.

So if I take s to be an infinitely small part of the wire, I can find the magnetic field is by taking the equation given above and integrating from neg. infinity to pos. infinity, because no length is given for the wire. I'm just going to assume it's an infinitely long wire.

I did that and I get (1.26E-6)I / (2pi(r))

Plugging in I and r I get the answer as 6.01E-5 T.

Is that right?
 
  • #4
rl.bhat
Homework Helper
4,433
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Yes. That is right.
 

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