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I was thinking about the magnetic field at the center of a loop of current and tried to calculate the field using results from Ampere's law. I understand that directly using Ampere's law does not work as there is not a suitably symmetric loop to consider for a current loop; however, I am struggling to see why my approach fails to circumvent this. I did the following calculation:

For a straight wire, we know that Ampere's law tells us that the magnetic field due to this wire (direction given by the right-hand-rule) is $$B = \frac{\mu_0 I}{2\pi r}$$ Now imagine taking a straight wire, and curling it into a circular loop of radius $R$. At the center, magnetic fields from each current element forming the loop add together. Using the calculation for straight wires, each current element of length $dl = Rd\theta$ in the loop gives $$dB = \frac{\mu_0 I}{2\pi R}R d\theta$$ Thus, integrating over the loop, the magnetic field at the center is $$B = \mu_0 I$$

Obviously this is incorrect -- the magnetic field at the center of a current loop is $$B= \frac{\mu_0 I}{2R}$$ per the Biot–Savart law. But WHY does this approach not work.

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  • $\begingroup$ Note that $B = \mu_0 I$ doesn’t even have the correct dimensions. $\endgroup$
    – Ghoster
    Mar 29 at 3:48
  • $\begingroup$ There is one obvious oversight, you have used the magnetic field for an infinitely long wire as the magnetic field of an infinitesimally small wire. Use the formula for a wire of finite length, replace $\sin \left(d\theta \right)$ with $d\theta $ multiply the magnetic field with the number of such infinitesimally small wires which is $\frac{2\pi }{2d\theta }$ you will get the same result. $\endgroup$ Mar 29 at 4:53
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    $\begingroup$ Curling a straight wire totally changes the symmetry character. It is more of a mystery why you think it would be a workable attempt. $\endgroup$ Mar 29 at 6:01

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Refer to my comment to as to why your approach is not entirely correct. $$ \begin{aligned} d B & =\frac{\mu_0 I}{4 \pi R}(\sin d \theta-\sin (-d \theta)) \\ & =\frac{\mu_0 I}{4 \pi R}(2 \sin d \theta) \end{aligned} $$ number of such elements $=\frac{2 \pi}{2 d \theta}=\frac{\pi}{d \theta}$ $$ \begin{aligned} B & =\frac{\pi}{d \theta}\left(\frac{\mu_0 I}{4 \pi R}(2 \sin d \theta)\right) \quad \sin d \theta \approx d \theta \\ & =\frac{\mu_0 I}{2 R} \end{aligned} $$ enter image description here

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The magnetic field due to an infinitesimal current carrying element $Idl$ at a point which is at a distance $r$ from it is given by the equation (from Biot Savart law) $$dB=\frac{\mu_o}{4π} \frac{Idlsin\theta}{r^2}$$

where $\theta$ is the angle made by line joining the current carrying element and the point, with the wire.

In the case of a wire bent to form a circle, $dl=rd\theta$.

Note that $sin\theta=1$ since the angle made between the current carrying element and the centre of the circle is $\frac{π}{2}$ for all elements of the circle.

$$\int dB=\int_{0}^{2π} \frac{\mu_o}{4π} \frac{Ird\theta}{r^2}$$

On solving, we get

$$B=\frac{\mu_oI}{2r}$$

We do not use Ampere's law to find the magnetic field at the centre of a current carrying loop of wire since the magnetic field is not symmetrical around the amperian loop.

Using Ampere's law is the wrong way to approach this and trying to integrate it will not yield you the right result.

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In your approach you have calculated the magnetic field of a superposition of all infinite currents carrying straight lines tangent to the loop. It is not surprising that this generates a different field than a loop.enter image description here

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