We are tasked with finding the current induced in the 110-turn coil based on the information provided. The solenoid creates a changing magnetic field, which induces a current in the coaxial coil due to Faraday's Law of Induction.

Known information:

1. Number of turns in the coil (N) = 110
2. Radius of the coil (r) = 2.8 cm = 0.028 m
3. Resistance of the coil (R) = 7.6 Ω
4. Turns per cm of the solenoid (n) = 320 turns/cm = 32000 turns/m
5. Diameter of the solenoid = 4.6 cm, which gives a radius of $r_s = 0.023 \, \text{m}$.
6. Change in current of the solenoid ($\Delta I_s$) = 1.3 A (from 1.3 A to 0 A)
7. Time interval ($\Delta t$) = 28 ms = 0.028 s

Step-by-step Solution:

1. Magnetic Field of the Solenoid: The magnetic field inside a solenoid is given by: $B_s = \mu_0 n I_s$ where:

   • $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T m/A}$),
   • $n$ is the number of turns per unit length,
   • $I_s$ is the current in the solenoid.

2. Magnetic Flux through the Coil: The magnetic flux $\Phi$ through the coil (with $N$ turns) is: $\Phi = N B_s A$ where $A$ is the area of the coil. Since the coil is coaxial with the solenoid, the flux only depends on the area of the solenoid. The area $A$ of the solenoid's cross-section is: $A_s = \pi r_s^2$

3. Induced EMF in the Coil: According to Faraday's Law, the induced EMF is the rate of change of magnetic flux: $\mathcal{E} = - N \frac{\Delta \Phi}{\Delta t} = - N \frac{\Delta (B_s A_s)}{\Delta t}$ Since $B_s = \mu_0 n I_s$, the change in magnetic flux $\Delta \Phi$ can be written as: $\Delta \Phi = N A_s \mu_0 n \Delta I_s$ Substituting into the equation for EMF: $\mathcal{E} = - N A_s \mu_0 n \frac{\Delta I_s}{\Delta t}$

4. Induced Current in the Coil: Using Ohm's Law, the induced current $I$ is related to the EMF by: $I = \frac{\mathcal{E}}{R}$ Substituting the expression for $\mathcal{E}$: $I = \frac{N A_s \mu_0 n \Delta I_s}{R \Delta t}$

5. Substitute the Known Values:

   • $N = 110$
   • $r_s = 0.023 \, \text{m}$
   • $A_s = \pi (0.023)^2 \, \text{m}^2$
   • $\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}$
   • $n = 32000 \, \text{turns/m}$
   • $\Delta I_s = 1.3 \, \text{A}$
   • $R = 7.6 \, \Omega$
   • $\Delta t = 0.028 \, \text{s}$

   Now, let's compute the induced current $I$.The induced current in the 110-turn coil is approximately 0.045 A (or 45 mA) during the given time interval of 28 ms.

Would you like more details or have any questions? Here are some follow-up questions for deeper understanding:

1. How would the induced current change if the solenoid's current dropped at a slower rate (i.e., over a longer time interval)?
2. What is the effect of increasing the resistance of the coil on the induced current?
3. How does the number of turns in the coil affect the induced EMF and current?
4. What would happen to the induced current if the solenoid had a larger radius?
5. How would the direction of the induced current change if the solenoid's current increased instead of decreased?

Tip: The faster the change in the magnetic field (i.e., the larger the $\Delta I / \Delta t$), the greater the induced EMF and current in the coil.