A long solenoid of diameter 0.1 m has \( 2 \times 10^{4} \) turns per meter. At the centre of the solenoid a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is \( 10\pi^{2}\,\Omega \), then the total charge flowing through the coil during this time is ___________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 32\,\mu\mathrm{C} \)
By Faraday's law the induced emf is \( e = -N\,d\phi/dt \), and the current is \( e/R \). The charge that flows is the current multiplied by the time, so the time cancels and only the total change of flux linkage matters.
\( \displaystyle q = \frac{N\,\Delta\phi}{R} \)
The coil sits inside the solenoid, where the field is uniform, \( B = \mu_0 n I \). Its own area (radius 0.01 m) is smaller than the solenoid's, so the flux through one turn is \( B \times \pi r^{2} \) with the coil's radius.
\( \displaystyle \Delta B = \mu_0 n\,\Delta I = 4\pi\times10^{-7} \times 2\times10^{4} \times 4 = 32\pi\times10^{-3}\,\mathrm{T} \)
\( \displaystyle A = \pi r^{2} = \pi \times (0.01)^{2} = \pi\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle N\,\Delta\phi = N\,\Delta B\,A = 100 \times 32\pi\times10^{-3} \times \pi\times10^{-4} = 32\pi^{2}\times10^{-5}\,\mathrm{Wb} \)
\( \displaystyle q = \frac{32\pi^{2}\times10^{-5}\,\mathrm{Wb}}{10\pi^{2}\,\Omega} = 3.2\times10^{-5}\,\mathrm{C} = 32\,\mu\mathrm{C} \)