Magnetic Field Current Loop

  • Thread starter jromega3
  • Start date
  • #1
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Homework Statement



A loop carries current I = 2.5 A in the x-y plane as shown in the figure above. The loop is made in the shape of a circular arc of radius R = 4 cm from qo = 60 ° to q = 360 ° . The loop is completed by horizontal and vertical sections as shown.
What is BA, the magnitude of the magnetic field at point A, the center of the circular arc?

Homework Equations



For the loop part....
loopc3.gif


The Attempt at a Solution



Well, that's it if it were a complete loop, but this is 300/360, or 5/6 of a loop. So I take that and divide by 1.2 and I get 3.27e^-5.
Not sure about the other ones. My calculus is rusty at best.

It's hard to visualize maybe, but it's basically a circle from the 60 degree above the horizon all the way till the end, with a straight line going down to the horizontal center line and then a line connecting that line to the 0 degree mark on the curve.

As always, any help is appreciated.
 

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Answers and Replies

  • #2
rl.bhat
Homework Helper
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Now you have to find the field due to the vertical section of the loop. By trig. you can see the length of the vertical section is 2sqrt3. The horizontal section does not produce any field at the centerIn the relevant equation take the integration of dx*sintheta/r^2 where dx is a small element of the straight vertical conductor, r is the distance of the element dx from the center and theta is the angle between the conductor and r.
 

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