Question

Suppose the current density changes slowly enough that we can (to good approximation) ignore all higher derivatives in the Taylor expansion

$\mathit{J}\left(t_r\right)\mathbf{=}\mathit{J}\left(t\right)\mathbf{+}\left(t_r-t\right)\mathit{J}\left(t\right)\mathbf{+}\mathbf{\dots }$

(for clarity, I suppress the r-dependence, which is not at issue). Show that a fortuitous cancellation in Eq. 10.38 yields

$\mathit{B}\left(r,t\right)\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\frac{\mathbf{J}\left(r\text{'},t\right)\mathbf{\times }\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}\mathcal{b}\mathbf{\text{'}}$.

That is: the Biot-Savart law holds, with J evaluated at the non-retarded time. This means that the quasistatic approximation is actually much better than we had any right to expect: the two errors involved (neglecting retardation and dropping the second term in Eq. 10.38 ) cancel one another, to first order.

Short answer

The fortuitous cancellation in Eq. 10.38 yields $B\left(r,t\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\int \left[\frac{J\left(r\text{'},t\right)\times \hat{r}}{r^2}\right]d\mathcal{b}\text{'}$.

Step-by-step solution

Expression for the current density and magnetic field strength.

Write the expression for the current density.

$\mathit{J}\left(t_r\right)\mathbf{=}\mathit{J}\left(t\right)\mathbf{+}\left(t_r-t\right)\mathbf{+}\mathit{J}\left(t\right)\mathbf{+}\mathbf{\dots }$

Here, J is the current density and$t_r$ is the retarded time.

Write the expression for the magnetic field strength.

$\mathit{B}\left(r,t\right)\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\left[\frac{J\left(r\text{'},t_r\right)}{r^2}+\frac{J\left(r,t_r\right)}{cr}\right]\mathbf{\times }\hat{\mathbf{r}}\mathit{d}\mathcal{b}\mathbf{\text{'}}$ …… (1)

Here, B is the magnetic field,${\mu }_0$ is the permeability of free space and c is the speed of light.

Prove B(r,t)=μ04π∫[Jr',t×r^r2]db' :

Write the expression for the retarded time.

$t_r=t-\frac{r}{c}$

Substitute the value of$t_r$ in equation (1).

$$\begin{gathered} B\left(r,t\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\int \left\{\left[\frac{J\left(r\text{'},t\right)}{r^2}+\frac{\left(t_r-t\right)J\left(t\right)}{r^2}\right]+\left[\frac{J\left(r\text{'},t_r\right)}{cr}\right]\right\}\times \hat{r}d\mathcal{b}\text{'} \\ B\left(r,t\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\int \frac{1}{r^2}\left[J\left(r\text{'},t\right)+\left(t_r-t\right)J\left(t\right)+\left(\frac{r}{c}\right)J\left(r\text{'},t_r\right)\right]\times \hat{r}d\mathcal{b}\text{'} \\ B\left(r,t\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\int \frac{1}{r^2}\left[J\left(r\text{'},t\right)-\frac{r}{c}J\left(r\text{'},t\right)+\frac{r}{c}J\left(r\text{'},t_{r1}\right)\right]\times \hat{r}d\mathcal{b}\text{'} \\ B\left(r,t\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\int \left[\frac{J\left(r\text{'},t\right)\times \hat{r}}{r^2}\right]d\mathcal{b}\text{'} \end{gathered}$$

Therefore, the fortuitous cancellation in Eq. 10.38 yields

$B\left(r,t\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\int \left[\frac{J\left(r\text{'},t\right)\times \hat{r}}{r^2}\right]d\mathcal{b}\text{'}$.