A long solenoid of radius R carries a time dependent current `I = I_(0) t(1 – t)`. A ring of radius 2R is placed coaxially near its centre. During the time interval`0 le t le 1`, the induced current `I_(R)` and the induced emf VR in the ring vary as:

Options

A. Direction of ` I _R ` remains unchanged and ` V _ R ` is maximum at ` t = 0.5 `
B. At t = 0.5 direction of ` I _ R ` reverses and ` V _ R ` is zero
C. At t =0.25 direction of ` I _ R ` reverses and ` V _ R ` is maximum
D. Direction of ` I _ R ` remains unchanged and ` V_R ` is zero at t = 0.25

Answer

To solve the problem, we need to analyze the induced current \( I_R \) and the induced emf \( V_R \) in a ring placed coaxially with a long solenoid carrying a time-dependent current. The current in the solenoid is given by:

\[ I(t) = I_0 t (1 - t) \]

### Step-by-Step Solution:

1. **Determine 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. For our case, substituting the expression for \( I(t) \):

\[ B(t) = \mu_0 n I_0 t (1 - t) \]

2. **Calculate the Magnetic Flux \( \Phi \) through the Ring**:
The area \( A \) of the ring of radius \( 2R \) is:

\[ A = \pi (2R)^2 = 4\pi R^2 \]

Therefore, the magnetic flux \( \Phi \) through the ring is:

\[ \Phi(t) = B(t) \cdot A = \mu_0 n I_0 t (1 - t) \cdot 4\pi R^2 \]

Simplifying, we have:

\[ \Phi(t) = 4\pi \mu_0 n I_0 R^2 t (1 - t) \]

3. **Find the Induced emf \( V_R \)**:
The induced emf \( V_R \) in the ring is given by Faraday's law of electromagnetic induction:

\[ V_R = -\frac{d\Phi}{dt} \]

Differentiating \( \Phi(t) \):

\[ V_R = -\frac{d}{dt} \left( 4\pi \mu_0 n I_0 R^2 t (1 - t) \right) \]

Using the product rule, we get:

\[ V_R = -4\pi \mu_0 n I_0 R^2 \left( (1 - t) + t(-1) \right) = -4\pi \mu_0 n I_0 R^2 (1 - 2t) \]

4. **Determine When \( V_R \) is Zero**:
Setting \( V_R = 0 \):

\[ 1 - 2t = 0 \implies t = 0.5 \]

5. **Determine When \( V_R \) is Maximum**:
The maximum value of \( V_R \) occurs at the endpoints of the interval or when the derivative of \( V_R \) with respect to \( t \) is zero. However, since \( V_R \) is linear in \( t \), we can check the values at \( t = 0 \) and \( t = 1 \):

- At \( t = 0 \):
\[ V_R(0) = -4\pi \mu_0 n I_0 R^2 (1 - 0) = -4\pi \mu_0 n I_0 R^2 \]

- At \( t = 1 \):
\[ V_R(1) = -4\pi \mu_0 n I_0 R^2 (1 - 2) = 4\pi \mu_0 n I_0 R^2 \]

Thus, \( V_R \) is maximum at \( t = 0 \) and \( t = 1 \).

6. **Determine the Induced Current \( I_R \)**:
The induced current \( I_R \) is related to the induced emf by:

\[ I_R = \frac{V_R}{R_{\text{ring}}} \]

where \( R_{\text{ring}} \) is the resistance of the ring. As \( V_R \) changes, \( I_R \) will also change accordingly.

### Summary of Results:
- \( V_R \) is maximum at \( t = 0 \) and \( t = 1 \).
- \( V_R = 0 \) at \( t = 0.5 \), indicating a reversal in the direction of the induced current \( I_R \).