A long solenoid of radius R carries a time (t) – dependent current `I=I_(0)(t-2t^(2))` A circular ring of radius = R is placed near the centre of the solenoid and plane of ring makes an angle `30^(@)` with the axis of solenoid. The time at which magnetic flux through the ring is maximum is :

Options

A. 0.5
B. 1.0 sec
C. 0.25 sec
D. 0.75 sec

Explanation

To solve the problem, we need to determine the time at which the magnetic flux through the circular ring is maximum. The magnetic flux is directly related to the magnetic field produced by the solenoid, which in turn depends on the current flowing through it.

### Step-by-step Solution:

1. **Identify the Current Function**:
The current flowing through the solenoid is given by:
\[ I(t) = I_0 (t - 2t^2) \]

2. **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 and \( \mu_0 \) is the permeability of free space. Since \( n \) and \( \mu_0 \) are constants, we can express the magnetic field as:
\[ B \propto I(t) \]

3. **Express the Magnetic Flux**:
The magnetic flux \( \Phi \) through the circular ring is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where \( A \) is the area of the ring and \( \theta \) is the angle between the magnetic field and the normal to the plane of the ring. Given \( \theta = 30^\circ \), we have:
\[ \Phi \propto I(t) \cdot A \cdot \cos(30^\circ) \]

4. **Maximize the Current**:
To find when the magnetic flux is maximum, we need to maximize the current \( I(t) \). We can do this by differentiating \( I(t) \) with respect to \( t \) and setting the derivative to zero:
\[ \frac{dI}{dt} = I_0 \left(1 - 4t\right) = 0 \]
Solving for \( t \):
\[ 1 - 4t = 0 \implies t = \frac{1}{4} \text{ seconds} \]

5. **Conclusion**:
The time at which the magnetic flux through the ring is maximum is:
\[ t = 0.25 \text{ seconds} \]