Inside a parallel plate capacitor the electric field E varies with times as `t^(2)`. The variation of inducted magnetic field with times is given by

Options

A. `t^(2)`
B. No variation
C. `t^(3)`
D. t

Explanation

To solve the problem of how the induced magnetic field varies with time inside a parallel plate capacitor where the electric field \( E \) varies as \( t^2 \), we can follow these steps:

### Step 1: Understand the relationship between electric field and charge in a capacitor.
The electric field \( E \) in a parallel plate capacitor is related to the charge \( Q \) on the plates and the capacitance \( C \) by the equation:
\[ Q = C \cdot V \]
where \( V \) is the potential difference across the plates. The capacitance \( C \) can be expressed as:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( A \) is the area of the plates and \( d \) is the separation between them.

### Step 2: Relate the electric field to the potential difference.
The electric field \( E \) is related to the potential difference \( V \) and the distance \( d \) between the plates by:
\[ E = \frac{V}{d} \]
If \( E \) varies as \( t^2 \), we can express \( V \) in terms of \( E \):
\[ V = E \cdot d \]

### Step 3: Determine the charge \( Q \) as a function of time.
Since \( E \propto t^2 \), we can write:
\[ E = k t^2 \]
for some constant \( k \). Therefore, the potential \( V \) becomes:
\[ V = k t^2 \cdot d \]
Substituting this into the charge equation:
\[ Q = C \cdot V = \frac{\varepsilon_0 A}{d} \cdot (k t^2 d) = \varepsilon_0 A k t^2 \]

### Step 4: Find the current \( I \) through the capacitor.
The current \( I \) is defined as the rate of change of charge with respect to time:
\[ I = \frac{dQ}{dt} = \frac{d}{dt}(\varepsilon_0 A k t^2) = 2 \varepsilon_0 A k t \]

### Step 5: Apply Ampere's Law to find the induced magnetic field \( B \).
According to Ampere's Law, the line integral of the magnetic field \( B \) around a closed loop is proportional to the current \( I \) passing through the loop:
\[ \oint B \cdot dl = \mu_0 I \]
For a circular loop of radius \( r \), the left side simplifies to \( B \cdot 2\pi r \):
\[ B \cdot 2\pi r = \mu_0 (2 \varepsilon_0 A k t) \]
Thus, we can solve for \( B \):
\[ B = \frac{\mu_0 (2 \varepsilon_0 A k t)}{2\pi r} = \frac{\mu_0 \varepsilon_0 A k t}{\pi r} \]

### Conclusion
The induced magnetic field \( B \) varies linearly with time \( t \):
\[ B \propto t \]