A capacitor with a capacitance of 100 F is connected to a 20 V battery for a long time before being disconnected. It is subsequently connected across a long solenoid that has 4000 turns per metre. It is observed that the potential difference across the capacitor drops to 90 % of its maximum value in 2.0 seconds. Determine the average magnetic field produced at the centre of the solenoid during this time interval.

Step-by-step solution

The initial charge on the capacitor is given by: Q_ i = C V_ i = 100 10⁻⁶ 20 = 2 10⁻³ C After 2.0 s , the potential difference drops to 90 % of its maximum value. The new potential difference is: V_ f = 0.9 20 = 18 V The charge on the capacitor at this time is: Q_ f = C V_ f = 100 10⁻⁶ 18 = 1.8 10⁻³ C The average current I flowing through the solenoid during this time interval t = 2.0 s is the rate of flow of charge: I = Q t = Q_ i - Q_ f t I = 2 10⁻³ - 1.8 10⁻³ 2.0 = 0.2 10⁻³ 2.0 = 10⁻⁴ A The magnetic field at the