Suppose the current density changes slowly enough that we can (to good approximation) ignore all higher derivatives in the Taylor expansion
J(tr)=J(t)+(tr−t)J˙(t)+⋯
(for clarity, I suppress the r-dependence, which is not at issue). Show that a fortuitous cancellation in Eq. 10.38 yields
B(r,t)=4πμ0∫i2J(r′,t)×i^dτ′.
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.