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Step-by-Step Solution
Step 1: Identify the Relevant Concept
This problem is based on electromagnetic induction. When the current in the solenoid changes, the magnetic flux through the smaller coil placed inside it also changes. This change in flux induces an emf in the coil, causing a charge to flow through it.
Step 2: Write Down the Known Quantities
Diameter of solenoid, $D = 0.1\,\text{m}$ (though radius is not directly needed for flux calculation through the coil)Turns per unit length of solenoid, $n = 2 \times 10^{4}\,\text{turns/m}$.
Coil inside solenoid has:
Number of turns, $N = 100$.
Radius, $r = 0.01\,\text{m}$.
Resistance of coil, $R = 10\,\pi^{2}\,\Omega$.
Initial current in the solenoid, $i_{1} = 4\,\text{A}$.
Final current in the solenoid, $i_{2} = 0\,\text{A}$.
Time in which current reduces, $\Delta t = 0.05\,\text{s}$ (not needed explicitly for total charge if rate is constant, but given for context).
Step 3: Determine the Magnetic Flux Through the Coil
The magnetic field $B$ inside a long solenoid, carrying current $i$, is given by:
$B = \mu_{0}\,n\,i$
where $ \mu_{0} $ is the permeability of free space ($4\pi \times 10^{-7}\,\text{H/m}$), and $n$ is the number of turns per unit length of the solenoid.
The flux $ \phi $ through one turn of the coil is:
$\phi = B \times A = \left(\mu_{0}\,n\,i\right) \times \left(\pi r^{2}\right)$
Hence, for $N$ turns of the coil, total flux linkage is $N \times \phi$.
Step 4: Calculate the Change in Flux
The initial flux linkage (when current is $i_{1}$) is:
$\phi_{i} = N \times \mu_{0}\,n\,i_{1}\,\pi r^{2}$
The final flux linkage (when current is $i_{2}$) is:
$\phi_{f} = N \times \mu_{0}\,n\,i_{2}\,\pi r^{2}$
Because the current goes from $4\,\text{A}$ to $0\,\text{A}$, the change in flux linkage is:
$\Delta \phi = \phi_{f} - \phi_{i} = N\,\mu_{0}\,n\,\pi r^{2} \left( i_{2} - i_{1} \right)$
Note that $ i_{2} - i_{1} = 0 - 4 = -4$, so the magnitude of change is positive but we keep track of signs as needed.
Step 5: Relate Induced Charge to Change in Flux
Total charge $q$ that flows through the coil when the current changes is given by the induced emf times the time (or more directly from Faraday's law integrated over time). The simplest expression is:
$q = \frac{\text{Change in flux linkage}}{R} = \frac{\Delta \phi}{R}
$
Thus, using the absolute value of the change in flux (since we focus on the net amount of charge), we get:
$q = \frac{N \,\mu_{0}\, n \,\pi r^{2} \left| i_{2} - i_{1} \right|}{R}.
$
Step 6: Substitute the Numerical Values
\(
\begin{aligned}
N &= 100, \\
r &= 0.01\,\text{m} \quad \Longrightarrow \quad \pi r^{2} = \pi \times (0.01)^{2} = \pi \times 10^{-4}, \\
\mu_{0} &= 4\pi \times 10^{-7}\,\text{H/m},\\
n &= 2 \times 10^{4}\,\text{turns/m},\\
(i_{1},\,i_{2}) &= (4,\,0)\,\text{A} \quad \Longrightarrow \quad |i_{2} - i_{1}| = 4, \\
R &= 10\,\pi^{2}\,\Omega.
\end{aligned}
\)
\(\displaystyle
q = \frac{100 \times (4\pi \times 10^{-7}) \times (2 \times 10^{4}) \times \left(\pi \times 10^{-4}\right) \times 4}{10\,\pi^{2}}.
\)
Step 7: Simplify Step by Step
Inside the numerator:
$ 4\pi \times 10^{-7} \times 2 \times 10^{4} = 8\pi \times 10^{-3} $
$ \pi \times 10^{-4} \times 4 = 4\pi \times 10^{-4} $
Combine: $ 8\pi \times 10^{-3} \times 4\pi \times 10^{-4} = 32\pi^{2} \times 10^{-7} $
Multiply by $100$: $ 100 \times 32\pi^{2} \times 10^{-7} = 3200 \pi^{2} \times 10^{-7} $
The denominator is $10 \pi^{2}$. Thus:
\(
q = \frac{3200 \pi^{2} \times 10^{-7}}{10 \pi^{2}} = \frac{3200 \times 10^{-7}}{10} = 320 \times 10^{-7} = 32 \times 10^{-6} \text{ C}.
\)
Step 8: Express the Final Answer
Thus,
\(
q = 32 \times 10^{-6}\,\text{C} = 32\,\mu\text{C}.
\)
Final Answer
32 μC
