long solenoid of radius R carries a time –dependent current `I(t) = I_(0) (12t - t^(3))` . A ring of radius 2R is placed coaxially near its middle. During the time interval ` 0 le t le 2 sqrt(3)` , the induced current `(I_(R))` and the induced EMF `(V_(R))` in the ring change as :

Options

A. At t = 1 direction of `I_(R)` reverses and `V_(R)` is maximum
B. At t = 2 direction of `I_(R)` reverses and is `V_(R)` zero
C. Direction of `I_(R)` remain unchanged and `V_(R)` is zero
D. Direction of `I_(R)` remains unchanged and `V_(R)` is maximum

Explanation

To solve the problem, we need to analyze the situation step by step. We have a long solenoid carrying a time-dependent current and a ring placed coaxially near its middle. We will calculate the induced EMF and the induced current in the ring.

### Step 1: Determine the expression for the current
The current in the solenoid is given by:
\[ I(t) = I_0 (12t - t^3) \]

### Step 2: Calculate the magnetic field inside the solenoid
The magnetic field \( B \) inside a long solenoid is given by:
\[ B = \mu_0 n I \]
where \( n \) is the number of turns per unit length of the solenoid and \( \mu_0 \) is the permeability of free space. Since the current \( I \) is time-dependent, we can express the magnetic field as:
\[ B(t) = \mu_0 n I(t) = \mu_0 n I_0 (12t - t^3) \]

### Step 3: Calculate the magnetic flux through the ring
The area \( A \) of the ring with radius \( 2R \) is:
\[ A = \pi (2R)^2 = 4\pi R^2 \]
Thus, the magnetic flux \( \Phi \) through the ring is:
\[ \Phi(t) = B(t) \cdot A = \mu_0 n I_0 (12t - t^3) \cdot 4\pi R^2 \]
\[ \Phi(t) = 4\pi \mu_0 n I_0 (12t - t^3) R^2 \]

### Step 4: Calculate the induced EMF
The induced EMF \( V_R \) in the ring is given by Faraday's law of electromagnetic induction:
\[ V_R = -\frac{d\Phi}{dt} \]
Calculating the derivative:
\[ \frac{d\Phi}{dt} = 4\pi \mu_0 n I_0 \left( 12 - 3t^2 \right) R^2 \]
Thus, the induced EMF becomes:
\[ V_R = -4\pi \mu_0 n I_0 (12 - 3t^2) R^2 \]

### Step 5: Calculate the induced current
The induced current \( I_R \) in the ring can be expressed as:
\[ I_R = \frac{V_R}{R_{\text{ring}}} \]
where \( R_{\text{ring}} \) is the resistance of the ring. Since \( R_{\text{ring}} \) is constant, the induced current will change as the induced EMF changes:
\[ I_R = -\frac{4\pi \mu_0 n I_0 (12 - 3t^2) R^2}{R_{\text{ring}}} \]

### Step 6: Analyze the behavior of \( V_R \) and \( I_R \)
1. **Finding when \( V_R = 0 \)**:
Set \( 12 - 3t^2 = 0 \):
\[ 3t^2 = 12 \]
\[ t^2 = 4 \]
\[ t = 2 \text{ seconds} \]

At \( t = 2 \) seconds, the induced EMF \( V_R \) becomes zero, indicating that the induced current \( I_R \) changes direction.

2. **Finding when \( V_R \) is maximum**:
The induced EMF is maximum at \( t = 0 \) because it starts from a maximum value and decreases as \( t \) increases.

### Conclusion
- The induced current \( I_R \) changes direction at \( t = 2 \) seconds.
- The induced EMF \( V_R \) is maximum at \( t = 0 \).