Question

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

• A. $ 16 \, \mu C $
• B. $32\, \mu C $
• C. $ 16 \, \pi \, \mu C $
• D. $32 \, \pi \, \mu C $

Answer 1

Answer: (D)

Given,
Number of turns, $n =100$
Radius, $r=0.01 m$
Resistance, $R =10 \pi^{2} \Omega$
As we know,
$ \epsilon=- N \frac{ d \phi}{ dt } $
$ =\frac{\epsilon}{ R }=-\frac{ N }{ R } \frac{ d \phi}{ dt }, \Delta I =-\frac{ N }{ R } \frac{ d \phi}{ dt } $
$ \frac{\Delta}{\Delta t }=-\frac{ N }{ R } \frac{\Delta \phi}{\Delta t } \Longrightarrow \Delta q =-\left[\frac{ N }{ R }\left(\frac{\Delta \phi}{\Delta t }\right)\right] \Delta t $
-ve sign shoes that induced emf opposes the change in flux. $ \Delta q =\frac{\mu_{0} n i \pi r ^{2}}{ R } $
$ \Delta q =\frac{4 \pi \times 10^{-7} \times 100 \times 4 \times \pi \times(0.01)^{2}}{10 \pi^{2}}=32 \mu C $