A long solenoid of diameter `0.1 m` has `2 xx 10^(4)` turns per metre. At the centre of the solenoid, a `100-`turns coil of radius `0.01 m` is placed with its axis coinciding with the constant rate from `+ 2 A` to `2 A` in `0.05 s`. Find the total charge (in `mu C`) flowing through the coil during this time when the resistance of the coil is `40 !=^(2) Omega`.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Mutual Inductance (M)
The formula for mutual inductance \( M \) between two coils is given by:
\[ M = \mu_0 n_1 n_2 A \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \) (permeability of free space)
- \( n_1 = 2 \times 10^4 \, \text{turns/m} \) (number of turns per meter of the solenoid)
- \( n_2 = 100 \, \text{turns} \) (number of turns in the coil)
- \( A = \pi r^2 \) (area of the coil)

Given the radius \( r = 0.01 \, \text{m} \):
\[ A = \pi (0.01)^2 = \pi \times 0.0001 \, \text{m}^2 = 0.0001\pi \, \text{m}^2 \]

Now substituting the values into the mutual inductance formula:
\[ M = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times (100) \times (0.0001\pi) \]
\[ M = 4\pi \times 10^{-7} \times 2 \times 10^4 \times 100 \times 0.0001\pi \]
\[ M = 8\pi^2 \times 10^{-5} \, \text{H} \]

### Step 2: Calculate the Induced EMF (E)
The induced EMF can be calculated using the formula:
\[ E = -M \frac{dI}{dt} \]
where:
- \( dI = I_f - I_i = 2 - (-2) = 4 \, \text{A} \) (the change in current)
- \( dt = 0.05 \, \text{s} \)

Substituting the values:
\[ E = - (8\pi^2 \times 10^{-5}) \frac{4}{0.05} \]
\[ E = - (8\pi^2 \times 10^{-5}) \times 80 \]
\[ E = -640\pi^2 \times 10^{-5} \, \text{V} \]

### Step 3: Calculate the Induced Current (I)
Using Ohm's law:
\[ I = \frac{E}{R} \]
where \( R = 40\pi^2 \, \Omega \):
\[ I = \frac{-640\pi^2 \times 10^{-5}}{40\pi^2} \]
\[ I = -16 \times 10^{-5} \, \text{A} = -0.00016 \, \text{A} \]

### Step 4: Calculate the Total Charge (Q)
The total charge flowing through the coil can be calculated using the formula:
\[ Q = I \times t \]
Substituting \( I = -0.00016 \, \text{A} \) and \( t = 0.05 \, \text{s} \):
\[ Q = -0.00016 \times 0.05 \]
\[ Q = -0.000008 \, \text{C} = -8 \, \mu C \]

### Final Answer
The total charge flowing through the coil during this time is \( 8 \, \mu C \).

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