Question

A capacitor of capacitance $100\mu F$ is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per metre. It is found that the potential difference across the capacitor drops to $90\%$ of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the centre of the solenoid during this period.

Answer 1

Given $C=100\mu F,V=20V$

$\therefore Q=CV$

$=1000\times {10}^{-6}\times 20$

$=2\times {10}^{-3}C$

Again $V^{\prime }=18V$

So, $Q^{\prime }=C{V}^{\prime }=1.8\times {10}^{-3}C$

Now current $i=\frac{Q-Q^{\prime }}{t}$

$=\frac{(2-1.8)\times {10}^{-3}}{2}$

$=\frac{2\times {10}^{-4}}{2}$

$=1\times {10}^{-4}A$

and $n=4000turns/m\left(Given\right)$

$\therefore B={\mu }_{o}ni$

$=4\pi \times {10}^{-7}\times 4000\times {10}^{-4}$

$=16\pi \times {10}^{-8}T.$