Q. 613.8( 4 Votes )

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.

Answer :

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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