Radius of a circular loop placed in a perpendicular uniform magnetic field is increasing at a constant rate of `r_(o)ms^(-1)` If at any instant radius of the loop is r, then emf induced in the loop at that instant will be

Options

A. `-2Brr_(0)`
B. `-2Bpir`
C. `-Bpirr_(0)`
D. `-2Bpirr_(0)`

Answer

To find the induced EMF in a circular loop whose radius is increasing at a constant rate in a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a circular loop with radius \( r \) placed in a uniform magnetic field \( B \) that is perpendicular to the plane of the loop. The radius of the loop is increasing at a constant rate of \( r_0 \) m/s.

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 circular loop, the area \( A \) is:
\[ A = \pi r^2 \]
Therefore, the magnetic flux becomes:
\[ \Phi = B \cdot \pi r^2 \]

3. **Applying Faraday's Law**:
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) in the loop is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]

4. **Differentiate the Magnetic Flux**:
We need to differentiate the magnetic flux \( \Phi \) with respect to time \( t \):
\[ \mathcal{E} = -\frac{d}{dt}(B \cdot \pi r^2) \]
Since \( B \) and \( \pi \) are constants, we can take them out of the differentiation:
\[ \mathcal{E} = -B \cdot \pi \cdot \frac{d}{dt}(r^2) \]

5. **Using the Chain Rule**:
Now, we apply the chain rule to differentiate \( r^2 \):
\[ \frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \]
Substituting this back into the equation for EMF:
\[ \mathcal{E} = -B \cdot \pi \cdot (2r \frac{dr}{dt}) \]

6. **Substituting the Rate of Change of Radius**:
We know that the radius is increasing at a constant rate of \( r_0 \) m/s, so \( \frac{dr}{dt} = r_0 \):
\[ \mathcal{E} = -B \cdot \pi \cdot (2r \cdot r_0) \]

7. **Final Expression for Induced EMF**:
Thus, the induced EMF in the loop at that instant is:
\[ \mathcal{E} = -2\pi B r r_0 \]

### Final Answer:
The induced EMF in the loop is:
\[ \mathcal{E} = -2\pi B r r_0 \]