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

The Correct Option is B

Sol. $\varepsilon=-N \frac{d \phi}{d t}$
$\left|\frac{\varepsilon}{R}\right|=\frac{N}{R} \frac{d \phi}{d t}$
$d q=\frac{N}{R} d \phi$
$\Delta Q=\frac{N(\Delta \phi)}{R}$
$\Delta Q=\frac{\Delta \phi_{\text {total }}}{R}$
$=\frac{(N B A)}{R}$
$=\frac{\mu_{0} n i \pi r^{2}}{R}$
Putting values
$=\frac{4 \pi \times 10^{-7} \times 100 \times 4 \times \pi \times(0.01)^{2}}{10 \pi^{2}}$
$\Delta Q=32\, \mu C$

Answer 2

The flowing current in the long solenoid will prompt emf in the inner coil. As the inner coil with some resistance, current flows through the inner coil. The total charge is given as the current flowing through the coil multiplied by the time for which the current is flowing.
Let us write down the given quantities.
The diameter of the long solenoid, d₁=0.1m=\(\frac{1}{10}\) Therefore the radius will be
r₁ = \(\frac{1}{2}\)×\(\frac{1}{10}\) = \(\frac{1}{20}\)m
The radius of the inner coil, r₂ = \(\frac{1}{100}\) m
Number of turns of the long solenoid, N₁=2×104
Number of turns of the inner coil, N₂ = 100
Resistance of the coil, R = 10𝜋²𝛺
 

Time, t=0.05 s Current’s rate of change is given as,
\(\frac{di}{dt}\) = \(\frac{(4-0)}{0.05}\) = 80 As^-1
The mutual inductance M on the second coil is given as:
M = \(\frac{(\mu _\theta N_1N_2A_2)}{l}\)

Here, μ₀ is a constant having value μ₀=4π×10^-7
A₂ is the area of the inner coil; A₂ = π×(r2)²
l is the length of the solenoid, l=1m as the number of turns is given per meter.
The induced emf e will be given as:

e = M \(\frac{di}{dt}\)
Substituting the values in the below equations of mutual inductance, we get
e = M×\(\frac{di}{dt}\) = \(\frac{4\pi \times 10^-7\times 2\times10^4\times100\times\pi\times(\frac{1}{100})}{1}\) × 80
⇒ e = 640×π²×10^-5 
We need to find the charge, the current is given as emf divided by resistance i = e/R and we know that i=\(\frac{e}{R}\) and we know that i=q/t therefore current will be given as:
q = \(\frac{e}{R}\)×t
Substituting the values of emf, e=640×π²×10^-5, resistance
R=10π²Ω and time t=0.05s,we get
q=\(\frac{(640\times\pi^2\times10^-5)}{10\pi^2\times0.05}\)
⇒q=64×10⁻⁵×5×10⁻²
⇒q=32×10⁻⁶C
⇒q=32μC
Therefore, the total charge flowing through the coil during the given time is 

q=32μC
Thus, option B is the correct option.