A cylindrical region of radius R contains a uniform magnetic field parallel to axis with magnitude that is changing linearly with time. If r is the radial distance of a point from axis of cylinder in a plane perpendicular to axis then the magnitude of the induced electrical field outside the cylinder is directly proportional to

Options

A. r
B. `r^2`
C. 1/r
D. `1//r^2`

Answer

To solve the problem, we need to determine how the induced electric field (E) outside a cylindrical region with a changing magnetic field is related to the radial distance (r) from the axis of the cylinder.

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a cylinder with a uniform magnetic field (B) parallel to its axis, which is changing linearly with time. We denote the radius of the cylinder as \( R \) and the radial distance from the axis as \( r \).

2. **Using Faraday's Law of Electromagnetic Induction**:
According to Faraday's law, the induced electromotive force (EMF) around a closed loop is equal to the negative rate of change of magnetic flux through the loop:
\[ \oint E \cdot dL = -\frac{d\Phi_B}{dt} \]
where \( \Phi_B \) is the magnetic flux.

3. **Calculating Magnetic Flux (\( \Phi_B \))**:
The magnetic flux through the area \( A \) of the cylinder is given by:
\[ \Phi_B = B \cdot A \cdot \cos(\theta) \]
Here, \( \theta = 0 \) degrees because the magnetic field is parallel to the axis of the cylinder, thus:
\[ \Phi_B = B \cdot \pi R^2 \]

4. **Finding the Rate of Change of Magnetic Flux**:
Since the magnetic field \( B \) is changing with time, we can express the rate of change of flux as:
\[ \frac{d\Phi_B}{dt} = \frac{dB}{dt} \cdot \pi R^2 \]

5. **Setting Up the Integral**:
The closed loop integral of the electric field around a circle of radius \( r \) (where \( r > R \)) is:
\[ \oint E \cdot dL = E \cdot (2\pi r) \]
Therefore, we can write:
\[ E \cdot (2\pi r) = -\frac{d\Phi_B}{dt} \]

6. **Substituting the Rate of Change of Flux**:
Substituting the expression for \( \frac{d\Phi_B}{dt} \) into the equation gives:
\[ E \cdot (2\pi r) = -\frac{dB}{dt} \cdot \pi R^2 \]

7. **Solving for the Electric Field (E)**:
Rearranging the equation to solve for \( E \):
\[ E = -\frac{R^2}{2r} \cdot \frac{dB}{dt} \]
The negative sign indicates the direction of the induced electric field, but we are interested in the magnitude.

8. **Final Relation**:
The magnitude of the induced electric field \( E \) is directly proportional to:
\[ E \propto \frac{1}{r} \]

### Conclusion:
Thus, the magnitude of the induced electric field outside the cylinder is directly proportional to \( \frac{1}{r} \).