Fall 2007 3 Em

A circular loop of wire is of radius R and carries current I. The wire lies in the plane z = 0 with its center at the origin of coordinates. Let ($\rho$,$\theta$,z) be the cylindrical coordinates.


(a) Magnetic field B at a given point (0, 0, z) on the z-axis.

(b) The radial component $B_{\rho} (\rho, \theta, z)$ of B at a distance $\rho \ll R$ off the z-axis.

Hint: For an arbitrary vector X

\begin{align} \mbox{div} \boldmath{X} = \frac{1}{\rho}\partial_{\rho} (\rho X_{\rho}) + \frac{1}{\rho} \partial_{\theta} X_{\theta} + \partial_z X_z \end{align}




(a) By symmetry, $B = B_z \hat{z}$ at $\rho = 0$. Let Idl be the differential current element along the ring, then the Biot-Savart law yields

\begin{align} dB_z = \frac{I \mbox{d}l}{c}\frac{\sin{\phi}}{R^2+z^2}, \sin{\phi} = \frac{R}{\sqrt{R^2+r^2}} \end{align}

where $\phi$ is the angle between the vector connecting this element to the observation point and the vertical. Integration over the ring leads to $\mbox{d}l \rightarrow 2\pi R$, and so

\begin{align} B_z = \frac{2\pi I }{c} \frac{R^2}{(R^2+z^2)^{3/2}} \end{align}

(b) By Taylor expansion

\begin{align} B_{\rho} \simeq \rho \partial_{\rho}B_{\rho}(0,0,z) \equiv \rho C, \; \rho \ll R \end{align}


\begin{align} 0 = \mbox{div} \bm{B}|_{\rho=0} = 2 C + \partial_z B_z \; \therefore \; C = - \partial_z B_z/ 2 \end{align}

Computing the last derivative, we finally get

\begin{align} B_{\rho} \simeq \frac{3 \pi I}{c} \frac{R^2 \rho z}{(R^2+z^2)^{5/2}} \end{align}
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