A capacitor of capacitance 100 μF 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 center of the solenoid during this period.
Given:
Capacitance(C) = 100 μF = 10-4 F
Initial Voltage of battery(V) = 20 V
Number of turns per meter of solenoid(n) = 4000
Final potential difference(V’) = 90% of 20 V = 18 V
Time taken(t) = 2 s
Formula used:
Initial charge stored in capacitor(Q) = CV,
where C = capacitance, V = potential difference
Therefore Q = (10-4 x 20) C = 2 x 10-3 C
New potential difference = V’ = 18 V
Therefore, New charge Q’ = CV’ = (10-4 x 18) = 1.8 x 10-3 C
Hence, current flowing in conductor(i) = (Q-Q’)/t, where Q = initial charge, Q’ = final charge, t = time taken
Therefore, i = 
= 10-4 A
Thus, average magnetic field at the center of solenoid = B = μ0ni,
where μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1, n = number of turns per unit length, i = current carried by the wire
Hence, from given data, B = (4π x 10-7 x 4000 x 10-4) T
= 0.5 x 10-6 T (Ans)
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