GREPhysics.NET: GR0177 #88

GR0177 #88

Problem

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Magnetism $\Rightarrow$ }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).

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.

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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.

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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

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

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sblmstyl
2008-10-15 21:57:49

Should't this question be organized with the other EM questions?

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fcarter
2008-10-13 17:07:58

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

John777
2011-11-08 06:44:02

Nice response!

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:

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

$r$ 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 $dI=Idl=IRd\theta$

$B=\frac{\mu _0}{4\pi}\int_0^\theta \frac{IRd\theta}{R^2}=\frac{\mu _0}{4\pi}\frac{I\theta}{R}$

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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.

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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.

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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.

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