A time varying uniform magnetic field passes through a circular region of radius R. The magnetic field is directed outwards and it is a function of radial distance 'r' and time 't' according to relation B-B_(0)rt. The induced electric field strength at a radial distance `R//2` from the centre will be.

Options

A. `(B_(0)R^(2))/(12)`
B. `(B_(0)R^(2))/(6)`
C. `(2B_(0)R^(2))/(3)`
D. `(B_(0)R^(2))/(16)`

Explanation

To solve the problem of finding the induced electric field strength at a radial distance \( R/2 \) from the center due to a time-varying magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Magnetic Field:**
The magnetic field \( B \) is given by the relation:
\[ B = B_0 \cdot r \cdot t \]
where \( B_0 \) is a constant, \( r \) is the radial distance, and \( t \) is time.

2. **Calculate the Magnetic Flux:**
The magnetic flux \( \Phi \) through a circular area of radius \( R/2 \) is calculated using the formula:
\[ \Phi = \int B \cdot dA \]
Since the magnetic field is uniform over the area and directed outwards, we can express the area element \( dA \) as \( dA = 2\pi r \, dr \). Therefore, the flux becomes:
\[ \Phi = \int_0^{R/2} B \cdot dA = \int_0^{R/2} B_0 \cdot r \cdot t \cdot 2\pi r \, dr \]

3. **Substituting the Magnetic Field:**
Substitute \( B \) into the flux equation:
\[ \Phi = \int_0^{R/2} B_0 \cdot r \cdot t \cdot 2\pi r \, dr = 2\pi B_0 t \int_0^{R/2} r^2 \, dr \]

4. **Calculating the Integral:**
The integral \( \int_0^{R/2} r^2 \, dr \) can be calculated as follows:
\[ \int_0^{R/2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^{R/2} = \frac{(R/2)^3}{3} = \frac{R^3}{24} \]

5. **Final Expression for Magnetic Flux:**
Substitute the result of the integral back into the flux equation:
\[ \Phi = 2\pi B_0 t \cdot \frac{R^3}{24} = \frac{\pi B_0 t R^3}{12} \]

6. **Using Faraday's Law of Induction:**
According to Faraday's law, the induced electromotive force (emf) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Therefore, we differentiate the flux with respect to time:
\[ \mathcal{E} = -\frac{d}{dt}\left(\frac{\pi B_0 t R^3}{12}\right) = -\frac{\pi B_0 R^3}{12} \]

7. **Relating Induced Electric Field to EMF:**
The induced electric field \( E \) around the circular path of radius \( R/2 \) is related to the emf by:
\[ \mathcal{E} = E \cdot 2\pi \left(\frac{R}{2}\right) \]
Thus,
\[ -\frac{\pi B_0 R^3}{12} = E \cdot \pi R \]

8. **Solving for the Induced Electric Field \( E \):**
Rearranging the equation gives:
\[ E = -\frac{B_0 R^2}{12} \]
Since we are interested in the magnitude, we can take the absolute value:
\[ E = \frac{B_0 R^2}{12} \]

### Final Answer:
The induced electric field strength at a radial distance \( R/2 \) from the center is:
\[ E = \frac{B_0 R^2}{12} \]