The way I have derived the magnetic field for inside the current loop of radius $R$ is very different from how I have seen it online. Generally they take the equation for the magnetic field at a point above the center of the loop, $\textbf{B}=\dfrac{\mu_0 IR^2}{2(R^2+z^2)^{3/2}}$, and let $z=0$. This looks like incorrect to me.

According to me the magnetic field would vary with the radius of the circle.

$\textbf{B}=I\oint\dfrac{d\ell \times (\vec{r_f}-\vec{r_s})}{|\vec{r_f}-\vec{r_s}|^3}$

I let $d\ell=-\sin\phi \ \hat{i} +\cos\phi \ \hat{j}$, $r_f=r\cos\phi \ \hat{i}+r\sin\phi \ \hat{j}$ and $r_s=R\cos\phi \ \hat{i}+R\sin\phi \ \hat{j}$. Which one is correct?


The formula that you quote from the online is correct if the curve is a circle, i.e. the radius is constant and one calculates the magnetic field in the middle of the circle.

Your formula uses the same assumption because you also took radius equal to $R$, i.e. $$r_s=Rcos\phi \ \hat{i} + Rsin\phi \ \hat{j},$$ but there are some mistakes:

1) $\text d \vec {\ell}$ is no correct, you forgot the element of length. The formula should be $$\text d \vec {\ell} = R \ \text d \phi (-sin\phi \ \hat{i} + cos\phi \ \hat{j}).$$ 2) As $\phi$ varies, the point at which you want to calculate the field, describes a circle. No, it's not correct. The Biot-Savart formula calculates the field at a certain point, due to the contribution of all the elements of the current. It is not true that each element of current contributes field to a different point. So you have to take $$r_f = r(cos\phi _0 \ \hat{i} + sin\phi _0\ \hat{j)},$$ i.e. let the point be fix.

However, since you choose your point $r_f$ not in the middle of the circle, the integration will be difficult. The integrand you get is

$$\frac {R}{[(Rcos\phi - rcos \phi_0)^2 + (Rsin\phi - rsin \phi_0)^2]^{3/2}} \begin {bmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {-sin\phi} & {cos\phi} & 0 \\ {Rcos\phi - rcos \phi_0} & {Rsin\phi - rsin \phi_0} & {0} \end {bmatrix} \text d \phi$$

$$ = -\frac {[R^2 - Rr \ cos(\phi - \phi_0)] \ \hat{k}}{[R^2 + r^2 -2Rr \ cos(\phi - \phi_0)]^{3/2}}\text d \phi$$

  • $\begingroup$ I think there is also a missing $\mu_0$ $\endgroup$ – Timaeus Feb 11 '15 at 5:55

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