A long solenoid of diameter $$0.1\,m$$  has $$2 \times {10^4}$$   turns per metre. 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 $$10\,{\pi ^2}\Omega ,$$   the total charge flowing through the coil during this time is

Solution :
Current induced in the coil is given by
$$\eqalign{ & i = \frac{1}{R}\left( {\frac{{d\phi }}{{dt}}} \right) \cr & \Rightarrow \frac{{\Delta q}}{{\Delta t}} = \frac{1}{R}\left( {\frac{{\Delta \phi }}{{\Delta t}}} \right) \cr} $$
Given, resistance of the solenoid,
$$R = 10\,{\pi ^2}\Omega $$
Radius of second and coil $$r = {10^{ - 2}}$$
$$\Delta t = 0.05s,\Delta i = 4 - 0 = 4\,A$$
Charge flowing through the coil is given by
$$\eqalign{ & \Delta q = \left( {\frac{{\Delta \phi }}{{\Delta t}}} \right)\frac{1}{R}\left( {\Delta t} \right) \cr & = {\mu _0}{N_1}{N_2}\pi {r^2}\left( {\frac{{\Delta i}}{{\Delta t}}} \right)\frac{1}{R}\Delta t \cr & = 4\pi \times {10^{ - 7}} \times 2 \times {10^4} \times 100 \times \pi \times {\left( {{{10}^{ - 2}}} \right)^2} \times \left( {\frac{4}{{0.05}}} \right) \times \frac{1}{{10{\pi ^2}}} \times 0.05 \cr & = 32 \times {10^{ - 6}}C = 32\,\mu C \cr} $$