Question

A capacitor of capacitance $100\mu F$ is connected to a battery of $20volts$ 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 $2.0seconds$. Estimate the average magnetic field produced at the centre of the solenoid during this period.

Answer 1

Given that,

$C=100\mu f,$

$V=20V,$

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

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

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

$t=2\sec$,

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

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

No. of turns per metre, $n=4000$

Thus, The average magnetic field at the centre of the solenoid is given by

$\therefore B={\mu }_{0}ni$

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

$=16\pi \times {10}^{-7}T$