A coil of wire having inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t=0, so that a time-dependent current `1_(1)(t)` starts following through the coil. If `I_(2)(t)` is the current induced in the ring, and B (t) is the magnetic field at the axis of the coil due to `I_(1)(t)` then as a function of time `(tgt0)`, the product `I_(2)(t)B(t)`

Options

A. increases with time
B. decreases with time
C. does not vary with time
D. passes through a mximum

Explanation

To solve the problem step by step, we will analyze the situation involving the coil, the induced current in the ring, and the magnetic field produced by the coil.

### Step 1: Understand the System
We have a coil with inductance \( L \) and resistance \( R \) connected to a battery at \( t = 0 \). A current \( I_1(t) \) starts flowing through the coil. Inside this coil, there is a conducting ring placed coaxially.

### Step 2: Determine the Magnetic Field \( B(t) \)
The magnetic field at the axis of the coil due to the current \( I_1(t) \) can be expressed as:
\[ B(t) = \mu_0 n I_1(t) \]
where \( \mu_0 \) is the permeability of free space and \( n \) is the number of turns per unit length of the coil.

### Step 3: Find the Induced EMF in the Ring
The induced electromotive force (EMF) \( E \) in the ring is due to the change in magnetic flux through it. The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B(t) \cdot A \]
where \( A \) is the area of the ring. The induced EMF can be expressed as:
\[ E = -\frac{d\Phi}{dt} = -A \frac{dB}{dt} \]
Since \( B(t) \) depends on \( I_1(t) \), we can write:
\[ E = -A \mu_0 n \frac{dI_1}{dt} \]

### Step 4: Find the Induced Current \( I_2(t) \)
The induced current \( I_2(t) \) in the ring can be found using Ohm's law:
\[ I_2(t) = \frac{E}{R} = -\frac{A \mu_0 n}{R} \frac{dI_1}{dt} \]
Let’s denote the constant \( k_1 = -\frac{A \mu_0 n}{R} \), then:
\[ I_2(t) = k_1 \frac{dI_1}{dt} \]

### Step 5: Determine \( I_1(t) \)
For an LR circuit, the current \( I_1(t) \) can be expressed as:
\[ I_1(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( \tau = \frac{L}{R} \) is the time constant.

### Step 6: Differentiate \( I_1(t) \)
Now, we differentiate \( I_1(t) \) to find \( \frac{dI_1}{dt} \):
\[ \frac{dI_1}{dt} = \frac{I_0}{\tau} e^{-\frac{t}{\tau}} \]

### Step 7: Substitute \( \frac{dI_1}{dt} \) into \( I_2(t) \)
Substituting \( \frac{dI_1}{dt} \) into the expression for \( I_2(t) \):
\[ I_2(t) = k_1 \cdot \frac{I_0}{\tau} e^{-\frac{t}{\tau}} \]

### Step 8: Find the Product \( I_2(t)B(t) \)
Now, we can find the product \( I_2(t)B(t) \):
\[ I_2(t)B(t) = \left(k_1 \cdot \frac{I_0}{\tau} e^{-\frac{t}{\tau}}\right) \left(\mu_0 n I_1(t)\right) \]
Substituting \( I_1(t) \):
\[ I_2(t)B(t) = k_1 \cdot \frac{I_0}{\tau} e^{-\frac{t}{\tau}} \cdot \mu_0 n \cdot I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]

### Step 9: Analyze the Behavior of \( I_2(t)B(t) \)
As \( t \) approaches 0, \( I_2(t)B(t) \) approaches 0. As \( t \) approaches infinity, \( I_2(t)B(t) \) also approaches 0. Therefore, there is a maximum value at some point in time.

### Conclusion
The product \( I_2(t)B(t) \) passes through a maximum value as time progresses.