Question

A long solenoid of diameter $0.1m$ has $2\times {10}^4$ turns per metre.At the centre of the solenoid, a coil of $100$ turns and radius $0.01m$ is placed with its axis coinciding with the solenoid axis.The current in the solenoid reduces at a constant rate to $0A$ from $4A$ in $0.05s$. If the resistance of the coil is $10{\pi }^2\Omega ,$ the total charge flowing through the coil during this time is.

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

Answer 1

Answer: (C)

$32\mu C$

Given,

Number of turns, $n=100$

Radius, $r=0.01m$

Resistance, $R=10{\pi }^2\Omega$

As we know,

$ϵ=-N\frac{d\varphi }{dt}$

$=\frac{ϵ}{R}=-\frac{N}{R}\frac{d\varphi }{dt}$, $\Delta I=-\frac{N}{R}\frac{d\varphi }{dt}$

$\frac{\Delta }{\Delta t}=-\frac{N}{R}\frac{\Delta \varphi }{\Delta t}⟹\Delta q=-\left[\frac{N}{R}\right(\frac{\Delta \varphi }{\Delta t}\left)\right]\Delta t$

$-$ve sign shoes that induced emf opposes the change in flux.

$\Delta q=\frac{{\mu }_{0}ni\pi r^2}{R}$

$\Delta q=\frac{4\pi \times {10}^{-7}\times 100\times 4\times \pi \times (0.01{)}^2}{10{\pi }^2}=32\mu C$