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GR9677 #69
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Problem
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This problem is still being typed. |
Electromagnetism }Ampere Law
Recall Ampere's Law,  .
Since the region one is interested in is a vacuum, one's Ampere Loop encloses all of the current. Thus, the field from each conductor is  = \mu_0 J \pi R^2) , where,  and R is the radius of the conductor. (This is a good approximation of the current, as one assumes that the vacuum region in the center is small compared to the area of the conductors.)
Making the approximation that  , one has  . Since both fields contribute in the center, the field is twice that,  , as in choice (A).
(Also, one can immediately eliminate all but choices (A) and (B) by the right-hand rule. One seeks a +y-direction field.)
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Alternate Solutions |
Richard
2007-10-01 18:35:55 | One should note that by geometry there will be no dependence on the radius r in the solution:
Thus (a) is the answer.
|  | sharpstones
2006-11-30 12:23:41 | It would be easier to not consider any vacuum in the center. Since you have current flowing in opposite directions they will naturally cancel each other out in the middle. Simple find the field at A due to one of the cylinders of current(it will be a distance d/2 from the center) and then multiply times 2. | |
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Comments |
The_Duck
2010-07-05 15:35:29 | If you can convince yourself that there is a magnetic field, and that the pi in pi r^2 cancels the pi in u_0/4pi, that alone gives you A (since B,C,D all have factors of pi. Also note that there's no answer with a factor of pi^2, so the pi's must cancel).
The_Duck
2010-07-05 15:41:02 |
Actually I think above my identification of the source of the pi's is wrong but the reasoning should still apply.
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|  | bucky0
2007-10-02 19:22:28 | Don't the fields cancel? If the currents are equal traveling in each direction, you should end up with the same B field contribution from each side, but in opposite directions, which would drop it to zero, correct?
antithesis
2007-10-04 20:48:53 |
But one is on the right of A, and one is to the left, so that they add up (think of right hand rule)
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| | Richard
2007-10-01 18:35:55 | One should note that by geometry there will be no dependence on the radius r in the solution:
Thus (a) is the answer.
ewhite2
2007-10-30 18:57:02 |
why?
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FortranMan
2008-08-29 12:47:04 |
because increasing the radius, while increasing the distance between their centers, would also increase the current, hence the answer should not be proportional to the inverse of r. Only d really matters in the magnitude of the field.
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hanin
2009-09-29 22:37:52 |
Why increasing radius would also increase current? I imagine this like 2 pipes carrying water; no matter how much you enlarge your your pipe, the same amount of water will always flow if there is no change in the source.
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dogsandfrogs
2009-10-08 12:24:51 |
Current is given as a current density, not a set value. Density is the same regardless of r, hence why the answer should not depend on r.
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| | sharpstones
2006-11-30 12:23:41 | It would be easier to not consider any vacuum in the center. Since you have current flowing in opposite directions they will naturally cancel each other out in the middle. Simple find the field at A due to one of the cylinders of current(it will be a distance d/2 from the center) and then multiply times 2. | | santon
2006-11-02 15:08:53 | | | yosun
2005-11-11 21:24:16 | keflavich: thanks for the typo-alert; it has been corrected. | | keflavich
2005-11-11 18:27:43 | Typo: "All but choices A and B because they have a +y field" should probably replace the last thing in parentheses. | |
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