A long solenoid of diameter 0.1 m has `2 xx 10^(4)` turns per meter. At centre of the solenoid is 100 turns coil of radius 0.01 m placed with its axis coinciding with solenoid axis. The current in the solenoid reduce at a constant rate to 0A 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

Options

A. `32pimuC`
B. `16muC`
C. `32muC`
D. `16pimuc`

Explanation

To solve the problem step by step, we will calculate the total charge flowing through the coil during the time the current in the solenoid decreases.

### Step 1: Identify the parameters
- Diameter of the solenoid = 0.1 m, hence radius \( r_s = 0.05 \) m.
- Turns per meter of the solenoid \( n_s = 2 \times 10^4 \) turns/m.
- Number of turns in the coil \( N = 100 \).
- Radius of the coil \( r_c = 0.01 \) m.
- Initial current \( I_i = 4 \) A, final current \( I_f = 0 \) A.
- Time duration \( t = 0.05 \) s.
- Resistance of the coil \( R = 10 \pi^2 \, \Omega \).

### Step 2: Calculate the change in current
The change in current \( \Delta I \) is given by:
\[ \Delta I = I_f - I_i = 0 - 4 = -4 \, \text{A} \]

### Step 3: Calculate the magnetic field inside the solenoid
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 n_s I \]
Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 4: Calculate the initial magnetic field
Using the initial current \( I_i = 4 \) A:
\[ B_i = \mu_0 n_s I_i = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 4 \]
Calculating this gives:
\[ B_i = 32\pi \times 10^{-3} \, \text{T} \]

### Step 5: Calculate the final magnetic field
Using the final current \( I_f = 0 \):
\[ B_f = \mu_0 n_s I_f = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 0 = 0 \, \text{T} \]

### Step 6: Calculate the change in magnetic field
\[ \Delta B = B_f - B_i = 0 - 32\pi \times 10^{-3} = -32\pi \times 10^{-3} \, \text{T} \]

### Step 7: Calculate the area of the coil
The area \( A \) of the coil is given by:
\[ A = \pi r_c^2 = \pi (0.01)^2 = \pi \times 10^{-4} \, \text{m}^2 \]

### Step 8: Calculate the change in magnetic flux
The change in magnetic flux \( \Delta \Phi \) through the coil is given by:
\[ \Delta \Phi = N \cdot A \cdot \Delta B \]
Substituting the values:
\[ \Delta \Phi = 100 \cdot \pi \times 10^{-4} \cdot (-32\pi \times 10^{-3}) \]
\[ \Delta \Phi = -3200\pi^2 \times 10^{-7} \, \text{Wb} \]

### Step 9: Calculate the induced EMF
Using Faraday's law, the induced EMF \( \mathcal{E} \) is:
\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} = \frac{3200\pi^2 \times 10^{-7}}{0.05} \]
Calculating this gives:
\[ \mathcal{E} = 64000\pi^2 \times 10^{-7} \, \text{V} \]

### Step 10: Calculate the total charge \( Q \)
Using Ohm's law, the total charge \( Q \) is given by:
\[ Q = \frac{\mathcal{E} \cdot \Delta t}{R} \]
Substituting the values:
\[ Q = \frac{64000\pi^2 \times 10^{-7} \cdot 0.05}{10\pi^2} \]
\[ Q = \frac{3200\pi^2 \times 10^{-7}}{10\pi^2} = 320 \times 10^{-6} \, \text{C} = 32 \, \mu\text{C} \]

### Final Answer
The total charge flowing through the coil during this time is \( 32 \, \mu\text{C} \).