A capacitor with a capacitance of 100 is connected to a 20 V...

The initial charge on the capacitor is given by:

$Q_{i} = C V_{i} = 100 \times 10^{-6} \times 20 = 2 \times 10^{-3} \text{ C}$

After $2.0 \text{ s}$, the potential difference drops to $90\%$ of its maximum value. The new potential difference is:

$V_{f} = 0.9 \times 20 = 18 \text{ V}$

The charge on the capacitor at this time is:

$Q_{f} = C V_{f} = 100 \times 10^{-6} \times 18 = 1.8 \times 10^{-3} \text{ C}$

The average current $\bar{I}$ flowing through the solenoid during this time interval $\Delta t = 2.0 \text{ s}$ is the rate of flow of charge:

$\bar{I} = \dfrac{\Delta Q}{\Delta t} = \dfrac{Q_{i} - Q_{f}}{\Delta t}$

$\bar{I} = \dfrac{2 \times 10^{-3} - 1.8 \times 10^{-3}}{2.0} = \dfrac{0.2 \times 10^{-3}}{2.0} = 10^{-4} \text{ A}$

The magnetic field at the centre of a long solenoid is $B = \mu_{0} n I$. Therefore, the average magnetic field $\bar{B}$ is:

$\bar{B} = \mu_{0} n \bar{I}$

Substituting the given values ($n = 4000 \text{ m}^{-1}$ and $\mu_{0} = 4\pi \times 10^{-7} \text{ T m/A}$):

$\bar{B} = (4\pi \times 10^{-7}) \times 4000 \times 10^{-4}$

$\bar{B} = 16\pi \times 10^{-8} = 1.6\pi \times 10^{-7} \text{ T}$