GR0177 #88
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Alternate Solutions |
2010-03-22 13:22:47 | Method of Elimination:
(A) Out since B field is obviously not zero.
(B), (D), (E) Out since they don't have correct units.
Ans: (C)
I found this the fastest way to attack this problem. |  | fcarter 2008-10-13 17:07:58 | Here is the fast way: It's all about limits. In the limit of theta approaching 2pi the circle becomes complete. The field of a circular loop is well known and you ought to have memorized it. The only answer that reduces properly in the limit is C.
hybridusmanus 2010-06-30 16:14:29 |
If we had 1/4 of a loop, then our B-field at P is * 
the fraction of the circle we have is: 
so the B-field at P is: , which is (C).
It is worth it to memorize the B-field of:
infinite wire
the center of a loop
inside a solenoid
inside a toroid
The magnetic field at a point of interest can be found using super position of all these geometries.
Here, we only have contribution from a fraction of a loop (circle), and not from the wires as they are in line with the point.
so we take the fraction of the loop length to a full circle and scale our magnetic field equation.
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Comments |
2010-03-22 13:22:47 | Method of Elimination:
(A) Out since B field is obviously not zero.
(B), (D), (E) Out since they don't have correct units.
Ans: (C)
I found this the fastest way to attack this problem.
ken 2010-07-13 17:50:49 |
i think this is the easiest way to approach this problem as well
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ncanna1 2010-09-25 12:14:50 |
(B) DOES have correct units. It only differs from (C) by a factor of .
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Dodobird 2010-11-01 10:48:45 |
B has correct units.
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aqme28 2010-11-09 18:38:57 |
B and C have the same units, so you can't discredit B based soley on this.
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zero 2014-09-26 06:46:30 |
(B) and (C) have the same units
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zero 2014-09-26 06:47:06 |
(B) and (C) have the same units
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|  | sblmstyl 2008-10-15 21:57:49 | Should't this question be organized with the other EM questions? |  | fcarter 2008-10-13 17:07:58 | Here is the fast way: It's all about limits. In the limit of theta approaching 2pi the circle becomes complete. The field of a circular loop is well known and you ought to have memorized it. The only answer that reduces properly in the limit is C.
lattes 2008-10-13 19:33:31 |
Great solution!
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Poop Loops 2008-11-02 00:52:51 |
I don't have it memorized and I can't find it anywhere. Help?
I know uI = B*2*pi*r for a wire, so B = uI/(2*pi*r).
But for answer C at Theta = 2*pi, B = uI/(2R).
So how do I get rid of that pi by itself?
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zaijings 2009-03-31 14:51:38 |
the center B field of circular loop:
B=UI/2R
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hybridusmanus 2010-06-30 16:14:29 |
If we had 1/4 of a loop, then our B-field at P is * 
the fraction of the circle we have is: 
so the B-field at P is: , which is (C).
It is worth it to memorize the B-field of:
infinite wire
the center of a loop
inside a solenoid
inside a toroid
The magnetic field at a point of interest can be found using super position of all these geometries.
Here, we only have contribution from a fraction of a loop (circle), and not from the wires as they are in line with the point.
so we take the fraction of the loop length to a full circle and scale our magnetic field equation.
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John777 2011-11-08 06:44:02 |
Nice response!
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yummyhat 2017-10-27 03:46:41 |
thanks hybrid
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|  | Richard 2007-10-29 11:45:27 | The Biot-Savart law for a 1d current:

is the distance from the point to the source.
As Yosun pointed out, the contributions due to the line currents are zero since is in the same direction as . To find the field due to the curved segment, notice that is always perpendicular to and integrate:

is a constant so 

|  | Richard 2007-10-29 11:22:31 | How about this:
It's pretty clear that the magnetic field should be inversely proportional to but proportional to .
Also, as you know from Ampere's law the field goes as not of an Amperian loop. This "means" there should be a in the denominator.
Choice (C) fits....not all that solid I know. |  | globalphysics 2006-11-05 20:33:05 | I agree with the statement from the user above. Just a small detail: In the description of subject material it should be "Biot-Savart Law" not "Biot-Savert". Small thing. |  | herrphysik 2006-10-01 18:54:42 | Sorry for being so picky, but you should have or . I think I speak for every test-taker who finds his or her way to your site when I say your efforts are much appreciated. Thanks Yosun. |  |
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