2.II.17G

Electromagnetism | Part IB, 2006

Derive from Maxwell's equations the Biot-Savart law

B(r)=μ04πVj(r)×(rr)rr3dV\mathbf{B}(\mathbf{r})=\frac{\mu_{0}}{4 \pi} \int_{V} \frac{\mathbf{j}\left(\mathbf{r}^{\prime}\right) \times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}} d V^{\prime}

giving the magnetic field B(r)\mathbf{B}(\mathbf{r}) produced by a steady current density j(r)\mathbf{j}(\mathbf{r}) that vanishes outside a bounded region VV.

[You may assume that the divergence of the magnetic vector potential is zero.]

A steady current density j(r)\mathbf{j}(\mathbf{r}) has the form j=(0,jϕ(r),0)\mathbf{j}=\left(0, j_{\phi}(\mathbf{r}), 0\right) in cylindrical polar coordinates (r,ϕ,z)(r, \phi, z) where

jϕ(r)={kr0rb,hzh0 otherwise ,j_{\phi}(\mathbf{r})= \begin{cases}k r & 0 \leqslant r \leqslant b, \quad-h \leqslant z \leqslant h \\ 0 & \text { otherwise },\end{cases}

and kk is a constant. Find the magnitude and direction of the magnetic field at the origin.

[ Hint :hhdz(r2+z2)3/2=2hr2(h2+r2)1/2]\left[\text { Hint }: \quad \int_{-h}^{h} \frac{d z}{\left(r^{2}+z^{2}\right)^{3 / 2}}=\frac{2 h}{r^{2}\left(h^{2}+r^{2}\right)^{1 / 2}}\right]

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