A long solenoid of radius R carries a time (t)-dependent current `I(t)= I_(0)t^(2) (1-t)`. A conducting ring of radius 3R is placed co-axially near its middle. During the time interva `0 le t le 1`, the induced current `(I_(R ))` in the ring varies as: [Take resistance of ring to be `R_(0)`]

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Explanation

To solve the problem, we need to find the induced current \( I_R \) in a conducting ring of radius \( 3R \) placed coaxially with a long solenoid carrying a time-dependent current \( I(t) = I_0 t^2 (1 - t) \). We will follow these steps:

### Step 1: Determine the Magnetic Field Inside the Solenoid
The magnetic field \( B \) inside a long solenoid is given by the formula:
\[ B = \mu_0 n I \]
where \( n \) is the number of turns per unit length and \( I \) is the current flowing through the solenoid. Since the current is time-dependent, we substitute \( I(t) \):
\[ B(t) = \mu_0 n I_0 t^2 (1 - t) \]

### Step 2: Calculate the Magnetic Flux Through the Ring
The area \( A \) of the conducting ring is given by:
\[ A = \pi (3R)^2 = 9\pi R^2 \]
The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi(t) = B(t) \cdot A = \mu_0 n I_0 t^2 (1 - t) \cdot 9\pi R^2 \]
Thus, the magnetic flux becomes:
\[ \Phi(t) = 9\pi \mu_0 n I_0 R^2 t^2 (1 - t) \]

### Step 3: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Calculating the derivative:
\[ \frac{d\Phi}{dt} = 9\pi \mu_0 n I_0 R^2 \frac{d}{dt}[t^2 (1 - t)] \]
Using the product rule:
\[ \frac{d}{dt}[t^2 (1 - t)] = 2t(1 - t) - t^2 = 2t - 3t^2 \]
Thus, we have:
\[ \mathcal{E} = -9\pi \mu_0 n I_0 R^2 (2t - 3t^2) \]

### Step 4: Calculate the Induced Current
Using Ohm's law, the induced current \( I_R \) in the ring can be calculated as:
\[ I_R = \frac{\mathcal{E}}{R_0} = \frac{-9\pi \mu_0 n I_0 R^2 (2t - 3t^2)}{R_0} \]
This simplifies to:
\[ I_R = -\frac{9\pi \mu_0 n I_0 R^2}{R_0} (2t - 3t^2) \]

### Step 5: Analyze the Current Function
The expression for the induced current \( I_R \) is:
\[ I_R = -k (2t - 3t^2) \]
where \( k = \frac{9\pi \mu_0 n I_0 R^2}{R_0} \). The current will be zero when:
\[ 2t - 3t^2 = 0 \implies t(2 - 3t) = 0 \]
This gives \( t = 0 \) and \( t = \frac{2}{3} \).

### Conclusion
The induced current \( I_R \) varies as a quadratic function of time, reaching zero at \( t = 0 \) and \( t = \frac{2}{3} \).