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I would like if someone could clarify this issue for me: When dealing with a current $I$ running in a loop with radius $R$ and looking for the magnetic field in the middle of the loop. By using Ampere's law, i know that the current $I$ runs through a loop with the same radius $R$ ,we get that: $$\int(B*dl) = \mu I $$

$$B= \mu I/2\pi R $$

and when using BIO-SAVART we get that $$dl*r = |dl||r|sin90$$ and therefore i get that: $$B = \mu I/2R$$

Which is not the same result as with ampere's law. I obviously miss something, maybe i can't use Ampere's law? Anyway, if someone could help me out here I would really appreciate it. Thanks.

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up vote 4 down vote accepted

The Biot Savart law is

$${\bf B} = \frac{\mu_0}{4\pi} \oint \frac{ I\, d{\bf l} \times {\bf r}}{|{\bf r}|^3}$$

In this case $d{\bf l} \times {\bf r} = dl\,|r|$ directed along the loop axis and integrating around the closed loop leads to a B-field magnitude $ B = \mu_0\, I/2R$ as you suggest.

However, I think there is a problem with your application of Ampere's law. This is that $$ \oint {\bf B} \cdot d{\bf l} = \mu_0 I\, ,$$ where I is the current enclosed by the closed loop around which you do the line integral on the LHS.

Usually, to apply Ampere's law, you choose a loop to integrate over that has either a constant B-field, and/or with a direction that is either parallel or perpendicular to $d{\bf l}$ (so that the scalar product and/or line integral are much simplified). What loop have you done your integral around? Is the B-field constant along this path? I don't think so...

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I have uploaded a picture, i hope it would help clear what i meant. so loop obviously has the same radius $R$ as the loop has – user3921 Jul 18 '14 at 18:20
Well yes, that's what I assumed you did. Now draw vectors representing the B-field at each point around your loop. They are not constant in magnitude and not parallel to the line elements; so you have not calculated the LHS of Ampere's law correctly. – Rob Jeffries Jul 18 '14 at 18:40
I don't get it..the magnitude of B is the same around the amperian loop (it's always the same distance) , and as far as the direction, even in the simplest case of infinte wire the direction is "moving" around the amperian loop. it is always "radial", but isn't that the same case here? – user3921 Jul 18 '14 at 19:08
Ok I see. Well what you have done with the analysis in the original posting is show that this isn't true. This isn't a simple case like the infinite wire; there isn't radial symmetry (if r=0 at the wire) and the field isn't parallel to the loop you have drawn. Yes, you are always at the same distance from 1 tiny piece of wire all the way around your loop; but what about the rest of the wire - can you see this is asymmetric? – Rob Jeffries Jul 18 '14 at 19:29

You probably misapplied Ampere's law. This law is usually used to find magnetic field only in special cases when the contour integral can be found as a function of single field value based on symmetry.

Magnetic field of a circular current loop is not so simple and Ampere's law cannot be easily used to find it. In such cases, the method of choice is to use the Biot-Savart law (integrate the contributions to the field due to elements of the circuit) or find vector potential as a function of position and then derive magnetic field from it.

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Your calculation of $B$ using Ampere's Law is not correct. The integration surface should have its area perpendicular to the current and should have one of the sides go parallel to $\vec{B}$. The choice of surface, when using that form of Ampere's Law, is usually a square or rectangle. That's why using it for a single loop does not work, because $\vec{B}$ is not uniform at the center of a single loop, meaning it does not point in the same direction along a line. Ampere's Law is very useful if you want to calculate $\vec{B}$ inside a solenoid, because it is approximately uniform in this case. See this example.

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I have uploaded a picture so I'll be sure i get you right. sa far as I can see, the current and the surface area are perpendicular, the magnetic field is not pointing at the same dirction x/y/z term, but it does always points in "radial" direction – user3921 Jul 18 '14 at 18:13
$\vec{B}$ that you drew only points up at that specific point. Above the plane of the loop it does not point up anymore. To use Ampere's Law, $\vec{B}$ should be uniform, i.e. be constant along a line. – Physico Jul 18 '14 at 18:53
Also, you want $\vec{B}$ to be parallel to your contour, which is not the case in your picture. That's why it usually makes sense to use a rectangle. – Physico Jul 18 '14 at 18:59
I don't get it..the magnitude of $B$ is the same around the amperian loop, and as far as the direction, even in the simplest case of infinte wire the direction is "moving" around the amperian loop. it is always "radial", but isn't that the same case here? – user3921 Jul 18 '14 at 19:07
Here is a picture of $\vec{B}$ from a single current loop. As you can see, no where it is uniform. Also, even the magnitude of $\vec{B}$ is not the same on the plane of the loop. – Physico Jul 18 '14 at 19:39

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