Answer
$B=3.09 \times 10^{-20} \mathrm{~T}$.
Work Step by Step
(a) Here, the enclosed electric flux is found by integrating
$
\Phi_E=\int_0^r E 2 \pi r d r=t(0.500 \mathrm{~V} / \mathrm{m} \cdot \mathrm{s})(2 \pi) \int_0^r\left(1-\frac{r}{R}\right) r d r=t \pi\left(\frac{1}{2} r^2-\frac{r^3}{3 R}\right)
$
with SI units understood. Then (after taking the derivative with respect to time) Eq. 32-3 leads to
$$
B(2 \pi r)=\varepsilon_0 \mu_0 \pi\left(\frac{1}{2} r^2-\frac{r^3}{3 R}\right) .
$$
For $r=0.0200 \mathrm{~m}$ and $R=0.0300 \mathrm{~m}$, this gives $B=3.09 \times 10^{-20} \mathrm{~T}$.
