GREPhysics.NET: GR8677 #65

GR8677 #65

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Electromagnetism $\Rightarrow$ }Biot-Savart Law

One either remembers the field of a circular current loop or one derives it from the Bigot-Savart Law,

$\vec{B} = \frac{\mu_0}{4\pi}\frac{I d\vec{l}\times\hat{r}}{r^2}.<br />$

One finds that $d\vec{l} \times \hat{r} = dl$ , as they're perpendicular. $dB = \frac{\mu_0 I dl}{4\pi r^2}$ , and the vertical components of the field cancel, thus there are only the horizontal components (parallel to the axis of the ring). $r$ is the directest distance from a differential element on the wire to a point on the axis, and $x$ is the horizontal distance (along the axis) from the center of the ring to that point. Thus, the integral becomes,

$\int \frac{\mu_0 I dl}{4\pi}\frac{b}{(x^2+b^2)^{3/2}} = \frac{\mu_0 I b}{2 }\frac{b}{(x^2+b^2)^{3/2}}, <br />$

where $\int dl=2\pi b$ .

So anyway, the field, at a point anywhere along the axis of the loop is $\vec{B}=\frac{\mu_0 I b}{2 }\frac{b}{(x^2+b^2)^{3/2}}$ .

At a point far, far, away, $x >> b$ , and thus $\vec{B}\approx \frac{\mu_0 I b}{2 }\frac{b}{x^3} = \frac{\mu_0 I b^2}{2 x^3}$ .

Set $x=c$ , where $c$ is the fixed coordinate of that point, to get that the field is proportional to $Ib^2$ .

Incidentally, defining the magnetic dipole moment to be $\mu=IA=I\pi b^2$ , one finds the field of a magnetic dipole to be $\frac{\mu \mu_0}{2\pi x^3}$ . The field far from a current loop is the same as the field of a magnetic dipole.

Alternate Solutions

ironman
2008-07-17 11:41:24

There is an even easier way to solve this problem. Far away from the current loop, the magnetic field can be considered to be that of a dipole. Recall that the dipole moment is defined by $\mu=iA$ , where $A=\pi b^{2}$ (the area of the current loop). So $\mu=i\pi b^{2}$ . One can quickly see the correct answer is (B)!

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Comments

student2008
2008-10-12 06:22:26

Actually, the solution using magnetic dipole moment is the best one, since there's nothing about the position of the fixed point in the conditions of the problem. It can even lie not on the axis of the loop.

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ironman
2008-07-17 11:41:24

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jax
2005-12-03 07:26:49

It makes more sense if you say that $dl \times hat{r} = dl sin \theta$ , and if you draw the diagram, $\theta = \frac{b}{r}$ so the Biot-Savart law becomes,

$dB =\frac{\mu_o I b dl}{4 \pi r^3}$ , and $B = \frac{\mu_o I b}{4 \pi r^3} \int dl$ where $\int dl = 2 \pi b$

When I first read this solution I was confused as to where the extra b in the numerator comes from. It comes from the sin term in the cross product.

Man this site needs a preview option, I've got my fingers crossed that my latex code is correct and my clarification actually makes sense...

This page might make more sense than the way I typed it...

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html#c3

jax
2005-12-03 07:27:54

ugh that's actually $sin \theta = \frac{b}{r}$ at the beginning. And I forgot how to do a hat on top of a letter in latex, I thought it was $\hat{r}$ but I guess I am wrong?

jax
2005-12-03 07:28:04

Oh wait my hat worked that time :)

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