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 maximum.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Magnetic Field \( B(t) \)
The magnetic field \( B(t) \) 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, \( n \) is the number of turns per unit length of the coil, and \( I_1(t) \) is the time-dependent current flowing through the coil.

### Step 2: Express the Current \( I_1(t) \)
For a coil with inductance and resistance connected to a battery, the current \( I_1(t) \) can be described by the equation:
\[ I_1(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( I_0 \) is the maximum current, and \( \tau \) is the time constant of the circuit.

### Step 3: Substitute \( I_1(t) \) into \( B(t) \)
Substituting \( I_1(t) \) into the expression for \( B(t) \):
\[ B(t) = \mu_0 n I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]

### Step 4: Calculate the Induced EMF \( E \)
The induced EMF \( E \) in the conducting ring is given by Faraday's law of electromagnetic induction:
\[ E = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux through the ring. The magnetic flux \( \Phi \) can be expressed as:
\[ \Phi = B(t) \cdot A \]
where \( A \) is the area of the ring. Thus,
\[ E = -A \frac{dB(t)}{dt} \]

### Step 5: Differentiate \( B(t) \)
To find \( \frac{dB(t)}{dt} \):
\[ \frac{dB(t)}{dt} = \mu_0 n I_0 \frac{d}{dt}\left(1 - e^{-\frac{t}{\tau}}\right) = \mu_0 n I_0 \cdot \frac{1}{\tau} e^{-\frac{t}{\tau}} \]

### Step 6: Substitute \( \frac{dB(t)}{dt} \) into the EMF Equation
Substituting back into the EMF expression:
\[ E = -A \cdot \mu_0 n I_0 \cdot \frac{1}{\tau} e^{-\frac{t}{\tau}} \]

### Step 7: Calculate the Induced Current \( I_2(t) \)
The induced current \( I_2(t) \) in the ring can be found using Ohm's law:
\[ I_2(t) = \frac{E}{R} \]
where \( R \) is the resistance of the ring. Thus,
\[ I_2(t) = -\frac{A \mu_0 n I_0}{R \tau} e^{-\frac{t}{\tau}} \]

### Step 8: Calculate the Product \( I_2(t) B(t) \)
Now, we can find the product \( I_2(t) B(t) \):
\[ I_2(t) B(t) = \left(-\frac{A \mu_0 n I_0}{R \tau} e^{-\frac{t}{\tau}}\right) \left(\mu_0 n I_0 \left(1 - e^{-\frac{t}{\tau}}\right)\right) \]
This simplifies to:
\[ I_2(t) B(t) = -\frac{A \mu_0^2 n^2 I_0^2}{R \tau} e^{-\frac{t}{\tau}} \left(1 - e^{-\frac{t}{\tau}}\right) \]

### Step 9: Analyze the Behavior of \( I_2(t) B(t) \)
As \( t \to 0 \), \( I_2(t) B(t) \) approaches 0. As \( t \to \infty \), \( I_2(t) B(t) \) also approaches 0. Therefore, \( I_2(t) B(t) \) will have a maximum value at some finite time \( t \).

### Conclusion
The product \( I_2(t) B(t) \) passes through a maximum value at some point in time and returns to zero as \( t \) approaches infinity.

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