A flexible wire loop in the shape of a circle has a radius that grows linearly with time. There is a magnetic field perpendicular to the plane of the loop that has a magnitude inversely proportional to the distance from the centre of the loop, `B(r ) prop (1)/(r )` How does the `emf` `E` vary with time?

Options

A. (a) `E prop t^(2)`
B. (b) `E prop t`
C. ( c) `E prop sqrt(t)`
D. (d) `E` is constant

Explanation

To solve the problem, we need to analyze how the electromotive force (emf) \( E \) varies with time as the radius of the circular wire loop changes and as the magnetic field behaves according to the given conditions.

### Step-by-step Solution:

1. **Understanding the Problem**:
- We have a circular loop with radius \( r \) that increases linearly with time, \( r(t) = ct \) where \( c \) is a constant.
- The magnetic field \( B \) is inversely proportional to the distance from the center of the loop, i.e., \( B(r) \propto \frac{1}{r} \).

2. **Magnetic Flux Calculation**:
- The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the loop. For a circle, the area \( A = \pi r^2 \).
- Thus, we can express the magnetic flux as:
\[ \Phi = B \cdot \pi r^2 \]

3. **Substituting for Magnetic Field**:
- Since \( B \propto \frac{1}{r} \), we can write:
\[ B = \frac{k}{r} \]
where \( k \) is a proportionality constant.
- Therefore, the magnetic flux becomes:
\[ \Phi = \frac{k}{r} \cdot \pi r^2 = k \pi r \]

4. **Finding the Induced EMF**:
- The induced emf \( E \) is given by Faraday's law of electromagnetic induction:
\[ E = -\frac{d\Phi}{dt} \]
- We need to differentiate the magnetic flux with respect to time:
\[ \Phi = k \pi r \implies \frac{d\Phi}{dt} = k \pi \frac{dr}{dt} \]

5. **Substituting for \( \frac{dr}{dt} \)**:
- Since \( r(t) = ct \), we have:
\[ \frac{dr}{dt} = c \]
- Therefore, substituting this into the equation for \( \frac{d\Phi}{dt} \):
\[ \frac{d\Phi}{dt} = k \pi c \]

6. **Final Expression for EMF**:
- Thus, the induced emf is:
\[ E = -\frac{d\Phi}{dt} = -k \pi c \]
- Since \( k \), \( \pi \), and \( c \) are constants, the induced emf \( E \) is also a constant.

### Conclusion:
The induced emf \( E \) does not vary with time; it remains constant.