A long solenoid of diameter 0.1 m has 2×104 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 10π2Ω, the total charge flowing through the coil during this time is

$Given n = 2 \times 10^{4}; I = 4 A$

Initially I = 0 A

$∴ B_{i} = 0 or ϕ_{i} = 0$

Finally, the magnetic field at the centre of the solenoid is given as

$\begin{array}{l} B_{f} = μ_{o} nI \\ B_{f} = 4 π \times 10^{- 7} \times 2 \times 10^{4} \times 4 \\ B_{f} = 32 π \times 10^{- 3} T \end{array}$

Final magnetic flux through the coil is given as

$ϕ_{f} = nBA = 100 \times 32 π \times 10^{- 3} \times π \times (0.01)^{2} ϕ_{f} = 32 π^{2} \times 10^{- 5} {Tm}^{2}$

Induced charge,

$q = \frac{| Δ ϕ |}{R} = \frac{|ϕ_{f} - ϕ_{i}|}{R} = \frac{32 π^{2} \times 10^{- 5}}{10 π^{2}} = 32 \times 10^{- 6} C = 32 μ C$

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