A long solenoid of diameter 0 . 1m has 2 × 10 ^ 4 turns per meter....

In the given case, the solenoid produces a changing magnetic field. This field is linked with the coil inside and hence, there is an induced emf in the coil.

We have, from the laws of EMI:

$ε = - N\dfrac{dϕ}{dt}$

\left|\dfrac{ε}{R}\right| = \dfrac{N}{R}\dfrac{dϕ}{dt}$</p><p>The ratio given above is the induced current in the circuit which can further be written as the rate of flow of charge per unit time.<br /><br />$dq = \dfrac{N}{R}dϕ$<br /><br />$\DeltaQ = \dfrac{N\left(\Deltaϕ\right)}{R}$<br /><br />$\DeltaQ = \dfrac{\Delta{ϕ}_{total}}{R}$<br /><br />$= \dfrac{\left(NBA\right)}{R}$<br /><br />$= \dfrac{N{\text{\mu}_{0} ni\text{\pi}{r}^{2}}{R}$<br /><br />Putting values<br /><br />$= \dfrac{100 \times \left(4\pi \times {10}^{- 7}\right) \times \left(2 \times {10}^{4}\right) \times 4 \times \left[{\pi \times ( 0.01 )}^{2}\right]}{10{\text{\pi}}^{2}}$<br /><br />$\DeltaQ = 32\mu C$