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

Explanation

To solve the problem, we need to find the time at which the magnetic flux through the circular ring placed near the center of the solenoid is maximum. Here are the steps to arrive at the solution:

### Step 1: Understand 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:
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length of the solenoid,
- \( I \) is the current flowing through the solenoid.

### Step 2: Substitute the Time-Dependent Current
The current \( I \) is given as:
\[ I = I_0 (t - 2t^2) \]
Substituting this into the magnetic field equation, we have:
\[ B = \mu_0 n I_0 (t - 2t^2) \]

### Step 3: Calculate the Magnetic Flux Through the Ring
The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where:
- \( A \) is the area of the ring,
- \( \theta \) is the angle between the magnetic field and the normal to the area of the ring.

Since the plane of the ring makes an angle of \( 30^\circ \) with the axis of the solenoid, the angle \( \theta \) between the magnetic field and the area vector of the ring is:
\[ \theta = 90^\circ - 30^\circ = 60^\circ \]
Thus, the flux becomes:
\[ \Phi = B \cdot A \cdot \cos(60^\circ) \]

### Step 4: Substitute the Values into the Flux Equation
The area \( A \) of the circular ring is given by:
\[ A = \pi R^2 \]
Now substituting \( B \) and \( A \) into the flux equation:
\[ \Phi = \mu_0 n I_0 (t - 2t^2) \cdot \pi R^2 \cdot \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we can simplify:
\[ \Phi = \frac{\mu_0 n I_0 \pi R^2}{2} (t - 2t^2) \]

### Step 5: Differentiate the Flux with Respect to Time
To find the time at which the flux is maximum, we differentiate \( \Phi \) with respect to \( t \) and set the derivative to zero:
\[ \frac{d\Phi}{dt} = \frac{\mu_0 n I_0 \pi R^2}{2} \frac{d}{dt}(t - 2t^2) = 0 \]
Calculating the derivative:
\[ \frac{d}{dt}(t - 2t^2) = 1 - 4t \]
Setting this equal to zero gives:
\[ 1 - 4t = 0 \implies t = \frac{1}{4} \text{ seconds} \]

### Conclusion
The time at which the magnetic flux through the ring is maximum is:
\[ t = \frac{1}{4} \text{ seconds} \]

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