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

Given `C = 100muF, V = 20V`
`Q= CV`
`=100xx10^(-6)xx20`
`=2xx10^(-3)C`
` Again `V 18V`
So, `Q = CV = 1.8xx10^(-3)C`
` Now current i= (Q-Q)/(t)`
`(2-1.8)xx10^(-3))/(2)`
`=(2xx10(-4))/(2)`
`=1xx10^(-4) A`
` and `n= 4000 turns//m (Given)`
` :.B = mu_(0)ni`
`=4pixx10^(-7)xx4000xx10^(-4)`
`=16pixx10^(-8)T`