Radius ol a circular luop 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. `-2 Bpir`
C. `- Bpir_(0)r`
D. `-2 Bpir_(0)r`

Answer

To solve the problem, we need to find the induced EMF in a circular loop whose radius is increasing at a constant rate in a uniform magnetic field. We will use Faraday's law of electromagnetic induction, which states that the induced EMF (ε) in a closed loop is equal to the negative rate of change of magnetic flux (Φ) through the loop.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the radius of the circular loop at any instant be \( r \).
- The area \( A \) of the circular loop is given by the formula:
\[ A = \pi r^2 \]
- The magnetic field \( B \) is uniform and perpendicular to the loop.

2. **Calculate the Magnetic Flux (Φ)**:
- The magnetic flux \( Φ \) through the loop is given by:
\[ Φ = B \cdot A = B \cdot (\pi r^2) \]

3. **Differentiate the Magnetic Flux with Respect to Time**:
- To find the induced EMF, we need to differentiate the magnetic flux with respect to time \( t \):
\[ \frac{dΦ}{dt} = \frac{d}{dt}(B \cdot \pi r^2) \]
- Since \( B \) is constant, we can take it out of the differentiation:
\[ \frac{dΦ}{dt} = B \cdot \pi \cdot \frac{d(r^2)}{dt} \]

4. **Apply the Chain Rule**:
- Using the chain rule, we differentiate \( r^2 \):
\[ \frac{d(r^2)}{dt} = 2r \frac{dr}{dt} \]
- Given that \( \frac{dr}{dt} = r_0 \) (the rate at which the radius is increasing), we can substitute this into our equation:
\[ \frac{d(r^2)}{dt} = 2r \cdot r_0 \]

5. **Substitute Back into the Flux Derivative**:
- Now substituting back into the expression for \( \frac{dΦ}{dt} \):
\[ \frac{dΦ}{dt} = B \cdot \pi \cdot (2r \cdot r_0) = 2\pi B r r_0 \]

6. **Calculate the Induced EMF (ε)**:
- According to Faraday's law:
\[ ε = -\frac{dΦ}{dt} \]
- Therefore, the induced EMF is:
\[ ε = -2\pi B r r_0 \]

### Final Answer:
The induced EMF in the loop at that instant is:
\[ ε = -2\pi B r r_0 \]