Problem 23.3

23.3. a)
We first find the emf. $\vert\epsilon\vert = \frac{d\Phi}{dt}$
$\vert\epsilon\vert = \frac{d(B A)}{dt}$
$\vert\epsilon\vert = .5 \mu_{0} n A \frac{dI}{dt}$
$\vert\epsilon\vert = .48 \times 10^{-3} V$
Now, to find the current in the ring, we use Ohm's law.
$I_{ring}= \frac{\epsilon}{R}$
$I_{ring} = \frac{.00048}{.0003} ~=~ 1.60 A$

b)
To find the magnetic field in the ring we plug in our value for the current.
$B_{ring} = \frac{\mu_{0}I}{2R}$
$B_{ring} = 2.01 \times 10^{-5} T$

c)
The field due to the coils points downward, and its increasing, so therefore the field due to the ring points upward.