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Magnetism}Biot-Savert Law

The Biot-Savart law is d\vec{B}=\frac{\mu_0 I}{4\pi}\frac{d\vec{l}\times\hat{r}}{r^2}.

One notes that the non-bent wires have dl || r, and thus their cross-products go to 0.

The only contribution to the field is from the curved segment. Its field is B=\frac{\mu_0 I}{4\pi}\frac{\theta R}{R^2}, where dl=\theta R is just the arclength. Simplifying, one gets choice (C).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
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. 		Alternate Solution - Unverified
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 \frac{1}{4} * B_{loop} = \frac{\mu i}{2 R}

the fraction of the circle we have is: \frac{arc length -of- our- loop}{arc length- of- full- loop}= \frac{\theta R}{2\pi R} = \frac{\theta}{2\pi}

so the B-field at P is: (\frac{\theta}{2\pi})( \frac{\mu i}{2 R}) = \frac{\theta \mu i}{4\pi R} , 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.


Alternate Solution - Unverified
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
ncanna1
2010-09-25 12:14:50		 (B) DOES have correct units. It only differs from (C) by a factor of \frac{1}{\pi}.
Dodobird
2010-11-01 10:48:45		 B has correct units.
aqme28
2010-11-09 18:38:57		 B and C have the same units, so you can't discredit B based soley on this.
zero
2014-09-26 06:46:30		 (B) and (C) have the same units
zero
2014-09-26 06:47:06		 (B) and (C) have the same units
Alternate Solution - Unverified
sblmstyl
2008-10-15 21:57:49		Should't this question be organized with the other EM questions?NEC
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!
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?
zaijings
2009-03-31 14:51:38		 the center B field of circular loop:

B=UI/2R
hybridusmanus
2010-06-30 16:14:29		 If we had 1/4 of a loop, then our B-field at P is \frac{1}{4} * B_{loop} = \frac{\mu i}{2 R}

the fraction of the circle we have is: \frac{arc length -of- our- loop}{arc length- of- full- loop}= \frac{\theta R}{2\pi R} = \frac{\theta}{2\pi}

so the B-field at P is: (\frac{\theta}{2\pi})( \frac{\mu i}{2 R}) = \frac{\theta \mu i}{4\pi R} , 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.


John777
2011-11-08 06:44:02		 Nice response!
yummyhat
2017-10-27 03:46:41		 thanks hybrid
Alternate Solution - Unverified
Richard
2007-10-29 11:45:27		The Biot-Savart law for a 1d current:

d\vec{B}=\frac{\mu _0}{4\pi}\frac{\vec{dI}\times \hat{r}}{r^2}

r is the distance from the point P to the source.
As Yosun pointed out, the contributions due to the line currents are zero since dI is in the same direction as dr. To find the field due to the curved segment, notice that r is always perpendicular to dI and integrate:
B=\frac{\mu _0}{4\pi}\int \frac{dI}{R^2}

I is a constant so dI=Idl=IRd\theta

B=\frac{\mu _0}{4\pi}\int_0^\theta \frac{IRd\theta}{R^2}=\frac{\mu _0}{4\pi}\frac{I\theta}{R}
NEC
Richard
2007-10-29 11:22:31		How about this:
It's pretty clear that the magnetic field should be inversely proportional to R but proportional to \theta.
Also, as you know from Ampere's law the field goes as 1/length not 1/area of an Amperian loop. This "means" there should be a 1/\pi in the denominator.
Choice (C) fits....not all that solid I know.		NEC
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.Typo Alert!
herrphysik
2006-10-01 18:54:42		Sorry for being so picky, but you should have dl=Rd\theta or l=R\theta. 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.Typo Alert!

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Sorry for being so picky, but you should have dl=Rd\theta or l=R\theta. 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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