A capacitor of capacitance `50 muF` is connected to a battery of `20` volts for a long time and then disconnected from it.It is now connected across a long solendoid having `8000` turns per minute.It is found that the potential difference across the capacitor drops to `90%` of its maximum value is `2.0` seconds.The average magnetic field produced at the centre of the solenoid during this period is `Npixx10^(-8) T`.Then find value of `N`.

Answer

A

Explanation

`i=(Deltaq)/(Deltat)=(50xx10^(6)(20-18))/2=50muC`
`B=mu_(0)ni=4pixx10^(-7)xx8000xx50xx10^(-5)`
`=16pixx10^(-8)`
Alternate sol.
Let `q_(0)` is intial charge on capacitor `&q` is charge supplied at `t=t`.
`i=i_(0)sinomegat`
`rArr (q_(0)-q)=q_(0)cos omega t`.
`q_(0)/Cxx90/100=(q_(0)cos omegat)/C`
`cosomegat=9/10`
`cos2omega=9/10`
`i_(av)=(i_(0)underset(0)overset(2sec)intsin omegat dt)/2`
`=i_(0)/(2omega)[cos omegat]_(0)^(2)`
`i_(0)/(2omega)[1-cos 2omega]`
`i/(2omega)xx1/10=i_(0)/(20omega)`
`B_(av)=mu_(0)ni_(av)`
`4pixx10^(-7)xx8000xxi_(0)/(20omega)`
`4pixx10^(-7)xx8000xxq_(0)/20`
`=(4pixx10^(-7)xx8000xx50xx10^(-6)xx20)/20`
`B_(av)=16pixx10^(-8)T`