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 meter. 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.

Explanation

There is some resistance of winding wire used to construct a solenoid. So it is a kind of RC discharging circuit. For a given capacitance of capacitor amount of charge on the capacitor is proportional to the potential difference applied between the plates of capacitor. Hence, when potential difference becomes `80%,` charge also becomes` 80% `of its initial value.
`q = q_(0)e^(-t//tau`
` V = V_(0) e^(-t//tau)`
Initial amount of charge : `q_(1) = CV = 200 xx 10^(-6) xx 100 = 0.02 C`
Charge after `0.2` seconds: ` q_(2) = 0.8 xx 0.02 = 0.016 C`
`Delta q = q_(1) - q_(2) = 0.02 - 0.016 = 0.004 C`
` Deltat = 0.2 s`
Average current `i_(av) = (Delta q)/(Delta t) = (0.004)/(0.2) = 0.02 A`
Average magnetic inside the solenoid :
` B_(av) = mu_(0) ni_(av) = 4pi xx 10^(-7) xx 5,000 xx 0.02 = 4pi xx 10^(-5) T`