GR0177 #88



Alternate Solutions 
20100322 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 20081013 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 20100630 16:14:29 
If we had 1/4 of a loop, then our Bfield at P is *
the fraction of the circle we have is:
so the Bfield at P is: , which is (C).
It is worth it to memorize the Bfield 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.

 

Comments 
20100322 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 20100713 17:50:49 
i think this is the easiest way to approach this problem as well

ncanna1 20100925 12:14:50 
(B) DOES have correct units. It only differs from (C) by a factor of .

Dodobird 20101101 10:48:45 
B has correct units.

aqme28 20101109 18:38:57 
B and C have the same units, so you can't discredit B based soley on this.

zero 20140926 06:46:30 
(B) and (C) have the same units

zero 20140926 06:47:06 
(B) and (C) have the same units

  sblmstyl 20081015 21:57:49  Should't this question be organized with the other EM questions?   fcarter 20081013 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 20081013 19:33:31 
Great solution!

Poop Loops 20081102 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?

zaijings 20090331 14:51:38 
the center B field of circular loop:
B=UI/2R

hybridusmanus 20100630 16:14:29 
If we had 1/4 of a loop, then our Bfield at P is *
the fraction of the circle we have is:
so the Bfield at P is: , which is (C).
It is worth it to memorize the Bfield 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.

John777 20111108 06:44:02 
Nice response!

yummyhat 20171027 03:46:41 
thanks hybrid

  Richard 20071029 11:45:27  The BiotSavart 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 20071029 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 20061105 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 "BiotSavart Law" not "BiotSavert". Small thing.   herrphysik 20061001 18:54:42  Sorry for being so picky, but you should have or . I think I speak for every testtaker who finds his or her way to your site when I say your efforts are much appreciated. Thanks Yosun.  

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