Question

A long solenoid of diameter 0.1 m has $2\times {10}^4$ turns per metre. At the center of the solenoid, a 100-turn coil of radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid is decreased at a constant rate from +2 A to - 2 A in 0.05 s. Find the total charge (in $\mu C$) flowing through the coil during this time when the resistance of the coil is $40{\pi }^2\Omega$.

Answer 1

Field at center of solenoid = $B={\mu }_0{N}_{l}i$ where, $N_l$-number of turns per unit length
Flux through the inner loop = $100\times \pi {0.01}^{2}B=100\times \pi {0.01}^2\times {\mu }_0{N}_{l}i$
emf induced = rate of change of flux =$\frac{△\varphi }{△t}=100\times \pi {0.01}^2\times {\mu }_0{N}_l\frac{△i}{△t}$
Current through loop = $i=\frac{emf}{R}=\frac{△\varphi }{△t}=100\times \pi {0.01}^2\times {\mu }_0{N}_l\frac{△i}{△t}\frac{1}{R}$
Total charge through the coil during this time=$i△t=\frac{△\varphi }{△t}=100\times \pi {0.01}^2\times {\mu }_0{N}_l△i\frac{1}{R}=7.89\times {10}^{-5}C$