Question

A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of $20\omega$ is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.

Answer 1

Here r =2 cm $=2\times {10}^{-2}m$

n =100 turns/cm

=1000 turns/m

i =5 A

$B={\mu }_{0}ni=4\pi \times {10}^{-7}\times 10000\times 5$

$=20\pi \times {10}^{-3}T$

$n^2=100$ turns, $R=20\omega$

r =1 cm $={10}^{-2}m.$

Flux linking per turn the second coil

$=B\pi r^2=B\pi \times {10}^{-4}$

$\varphi$ (Total flux linking)

$=B{n}_2\pi r^2$

$=100\times \pi \times {10}^{-4}\times 20\pi \times {10}^{-3}$

Where current is reversed

$d\varphi ={\varphi }_2-{\varphi }_1$

$=2\times 100\times \pi \times {10}^{-4}\times 20\pi \times {10}^{-3}$

$e=\frac{-d\varphi }{dt}=\frac{4{\pi }^2\times {10}^{-4}}{dt}$

$I=\frac{e}{R}=\frac{4{\pi }^2\times {10}^{-4}}{dt\times 20}$

$q=Idt=\frac{4{\pi }^2\times {10}^{-4}}{20\times dt}\times dt$

$=2\times {10}^{-4}C.$