The diagram shows a solenoid carrying time varying current `I=I_0t` (`I_0` is constant) on the axis of this solenoid a conducting ring is being placed as shown in the figure. The mutual inductance of the ring and the solenoid is M and self inductance of the ring is L . if the resistance of the ring is R then the maximum current which can flow through the ring is

Options

A. `(2M+L)(I_0)/R`
B. `(MI_0)/R`
C. `(2M-L)(I_0)/R`
D. `(M+L)(I_0)/R`

Answer

To solve the problem, we need to find the maximum current \( I_{\text{max}} \) that can flow through the conducting ring placed on the axis of the solenoid carrying a time-varying current \( I = I_0 t \).

### Step-by-Step Solution:

1. **Understanding the Current in the Solenoid**:
The current in the solenoid is given by:
\[ I(t) = I_0 t \]
where \( I_0 \) is a constant.

2. **Finding the Magnetic Flux**:
The magnetic flux \( \Phi \) through the ring due to the solenoid can be expressed as:
\[ \Phi = M I(t) \]
where \( M \) is the mutual inductance between the solenoid and the ring.

3. **Substituting the Current into the Flux Equation**:
Substitute \( I(t) \) into the flux equation:
\[ \Phi = M (I_0 t) \]

4. **Calculating the Induced EMF**:
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) in the ring is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Substituting the expression for \( \Phi \):
\[ \mathcal{E} = -\frac{d}{dt}(M I_0 t) = -M I_0 \]

5. **Applying Ohm's Law**:
According to Ohm's law, the induced EMF is also equal to the product of the current \( I \) flowing through the ring and its resistance \( R \):
\[ \mathcal{E} = I R \]

6. **Setting the Two Expressions for EMF Equal**:
Equating the two expressions for EMF:
\[ -M I_0 = I R \]
Since we are interested in the magnitude, we can drop the negative sign:
\[ M I_0 = I R \]

7. **Solving for the Current \( I \)**:
Rearranging the equation to find the current \( I \):
\[ I = \frac{M I_0}{R} \]

8. **Identifying the Maximum Current**:
Since the current \( I \) is directly proportional to \( I_0 \) and the mutual inductance \( M \), the maximum current \( I_{\text{max}} \) that can flow through the ring is:
\[ I_{\text{max}} = \frac{M I_0}{R} \]

### Final Answer:
The maximum current which can flow through the ring is:
\[ I_{\text{max}} = \frac{M I_0}{R} \]