A solenoid has a cross sectional area of `6.0xx10^(-4) m^(2)`, consists of `400` turns per meter, and carries a current of `0.40 A`.`A 10` turn coil is wrapped tightly around the circumference of the solenoid.The ends of the coil are connected to a `1.5 Omega` resistor.Suddenly, a switch is opened and the current in the solenoid dies to zero in a time `0.050 s`,Find the average current in the coil during the time.

Explanation

To solve the problem, we need to find the average current induced in the coil when the current in the solenoid decreases to zero. We can use Faraday's law of electromagnetic induction and Ohm's law for this purpose.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Cross-sectional area of the solenoid, \( A = 6.0 \times 10^{-4} \, \text{m}^2 \)
- Number of turns per meter in the solenoid, \( n_s = 400 \, \text{turns/m} \)
- Current in the solenoid, \( I = 0.40 \, \text{A} \)
- Number of turns in the coil, \( n_c = 10 \)
- Resistance of the coil, \( R = 1.5 \, \Omega \)
- Time for the current to die to zero, \( \Delta t = 0.050 \, \text{s} \)

2. **Calculate the Magnetic Field in the Solenoid:**
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 n_s I \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space).
\[ B = (4\pi \times 10^{-7}) \times 400 \times 0.40 \]

3. **Calculate the Change in Magnetic Field:**
Since the current goes from \( 0.40 \, \text{A} \) to \( 0 \, \text{A} \), the change in magnetic field \( \Delta B \) is:
\[ \Delta B = B_{\text{final}} - B_{\text{initial}} = 0 - B \]

4. **Calculate the Change in Magnetic Flux:**
The magnetic flux \( \Phi \) through the coil is given by:
\[ \Phi = B \cdot A \]
Therefore, the change in flux \( \Delta \Phi \) is:
\[ \Delta \Phi = A \cdot \Delta B \]

5. **Calculate the Induced EMF:**
According to Faraday's law, the induced EMF \( \mathcal{E} \) in the coil is:
\[ \mathcal{E} = -n_c \frac{d\Phi}{dt} \]
Substituting \( \Delta \Phi \) and \( \Delta t \):
\[ \mathcal{E} = -n_c \frac{\Delta \Phi}{\Delta t} \]

6. **Calculate the Average Current in the Coil:**
Using Ohm's law, the average current \( I_c \) in the coil is:
\[ I_c = \frac{\mathcal{E}}{R} \]

7. **Substituting Values:**
Substitute all the calculated values into the equations to find \( I_c \).

### Final Calculation:
After performing the calculations, we find:
\[ I_c = 1.6 \times 10^{-5} \, \text{A} \]

### Summary:
The average current in the coil during the time when the solenoid's current dies to zero is \( 1.6 \times 10^{-5} \, \text{A} \).