Problem 23.17

23.17.
Taking one derivative of the field, we have that

Also, we can find the emf due to the changing B field.
$\vert\epsilon\vert = \frac{d\Phi}{dt}$
So, $\vert\epsilon\vert = A \frac{dB}{dt}$
$\vert\epsilon\vert = (\pi r_{1}^{2})(.06 t)$
We know that
$\vert\epsilon\vert = \oint \vec{E} \cdot d\vec{s}$
So, if we evaluate this along a closed circle, we have that
$E = \frac{\epsilon}{2\pi r}$
Plugging in, we have
$E = (\pi r_{1}^{2})(\frac{.06 t}{2 \pi r_{1}})$
At time, t = 3.00 s,
$E = 1.80 \times 10^{-3} N/C$
The electric field is perpendicular to r₁ and counterclockwise.