A long solenoid of radius `2cm` has `100 (turns) / (cm)` and carries a current of `5A`. A coil of radius `1 cm` having 100 tums and a total resistance of `20 ohm` is placed inside the solenoid coraxially, The coil is cornected to a galwanometer and the current in the solenoid is reversed in direction. If the charge flown through the galvanometer is `k times 10^(-oo) C`, then find `k`. (Take `pi^(2)=10` )

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the magnetic field inside the solenoid.
The magnetic field \( B \) inside a long solenoid is given by the formula:
\[ B = \mu_0 n I \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space),
- \( n \) is the number of turns per unit length,
- \( I \) is the current in the solenoid.

Given:
- The solenoid has \( 100 \, \text{turns/cm} \), which is equivalent to \( 10000 \, \text{turns/m} \) (since \( 1 \, \text{cm} = 0.01 \, \text{m} \)).
- The current \( I = 5 \, \text{A} \).

Substituting these values:
\[ B = (4\pi \times 10^{-7}) \times (10000) \times (5) \]

### Step 2: Simplify the expression for \( B \).
Calculating:
\[ B = 4\pi \times 10^{-7} \times 10000 \times 5 = 4\pi \times 5 \times 10^{-3} = 20\pi \times 10^{-3} \, \text{T} \]

Using \( \pi^2 = 10 \), we can approximate \( \pi \approx 3.16 \):
\[ B \approx 20 \times 3.16 \times 10^{-3} \approx 63.2 \times 10^{-3} \, \text{T} = 0.0632 \, \text{T} \]

### Step 3: Calculate the change in magnetic flux through the coil.
The magnetic flux \( \Phi \) through one turn of the coil is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the coil:
\[ A = \pi r^2 \]
Given the radius of the coil \( r = 1 \, \text{cm} = 0.01 \, \text{m} \):
\[ A = \pi (0.01)^2 = \pi \times 10^{-4} \, \text{m}^2 \]

Thus, the flux through one turn of the coil is:
\[ \Phi = B \cdot A = (0.0632) \cdot (\pi \times 10^{-4}) \]

### Step 4: Calculate the total flux linkage for the coil.
The total flux linkage \( \Psi \) for \( N \) turns is:
\[ \Psi = N \Phi = 100 \cdot (0.0632 \cdot \pi \times 10^{-4}) = 6.32 \cdot \pi \times 10^{-4} \, \text{Wb} \]

### Step 5: Calculate the induced EMF when the current is reversed.
The induced EMF \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{d\Psi}{dt} \]
Assuming the current is reversed instantaneously, the change in flux is \( 2\Psi \):
\[ \mathcal{E} = \frac{2 \cdot 6.32 \cdot \pi \times 10^{-4}}{dt} \]

### Step 6: Calculate the charge \( Q \) that flows through the galvanometer.
The charge \( Q \) can be calculated using:
\[ Q = \frac{\mathcal{E} \cdot t}{R} \]
where \( R = 20 \, \Omega \) is the resistance of the coil.

Assuming \( dt = 1 \, \text{s} \) for simplicity:
\[ Q = \frac{2 \cdot 6.32 \cdot \pi \times 10^{-4}}{20} \]

### Step 7: Substitute \( \pi \) and calculate \( k \).
Using \( \pi \approx 3.16 \):
\[ Q = \frac{2 \cdot 6.32 \cdot 3.16 \times 10^{-4}}{20} = \frac{39.94 \times 10^{-4}}{20} = 1.997 \times 10^{-4} \, \text{C} \]

### Step 8: Express \( Q \) in the form \( k \times 10^{-8} \).
\[ Q \approx 2 \times 10^{-4} \, \text{C} = k \times 10^{-8} \, \text{C} \]
Thus, \( k = 20 \).

### Final Answer:
The value of \( k \) is \( 2 \).