Problem 24.1

24.1.
From equation 24.2, we know that
$\oint \vec{B} \cdot d\vec{\ell} ~=~ \mu_{0} I + \mu_{0}\epsilon_{0}\frac{d\Phi_{E}}{dt}$
But, the current I = 0, so that term drops out. Also,
$\mu_{0}\epsilon_{0} = \frac{1}{c^{2}}$
So, we are left with
$\oint \vec{B} \cdot d\vec{\ell} ~=~ \frac{1}{c^{2}} \frac{d\Phi_{E}}{dt}$
Now, we know that
$\Phi_{E} = \int \vec{E} \cdot d\vec{\ell}$
$\Phi_{E} = E A$
Consequently, $\frac{d\Phi_{E}}{dt} = \frac{d}{dt} (E A)$
$\frac{d\Phi_{E}}{dt} = A \frac{dE}{dt}$
And the area through which the flux is flowing is
$A = \pi (\frac{d}{2})^{2}$
On the other side of the equation,
$\oint \vec{B} \cdot d\vec{\ell} ~=~ B (2\pi R)$
So, putting things together, we have
$(B)(2\pi R) = \frac{\pi d^{2}}{4c^{2}} \frac{dE}{dt}$
$B = \frac{d^{2}}{8c^{2}R} \frac{dE}{dt}$
Plug in the numbers, we have
$B = \frac{(.10 m)^{2}}{8 (3 \times 10^{8} m/s)^{2} (0.15 m)} (20 \frac{V}{m \cdot s})$
$B = 1.85 \times 10^{-18} T$
 
As far as the direction, the direction of the increase of the electric field is out of the paper. This will cause a magnetic field counterclockwise around the electric field. So, at point P, the magnetic field is directed upward.