A conducting square loop is placed in a magnetic field `B` with its plane perpendicular to the field. Now the sides of the loop start shrinking at a constant rate `alpha.` the induced emf in the loop at an instant when its side is a is

Options

A. `2aalphaB`
B. `a^2alphaB`
C. `2a^2alphaB`
D. `aalphaB`

Explanation

To solve the problem of finding the induced EMF in a conducting square loop placed in a magnetic field \( B \) while its sides are shrinking at a constant rate \( \alpha \), we can follow these steps:

### Step 1: Understand the relationship between side length and time
Let the initial side length of the square loop be \( l_i \). As the sides of the loop shrink at a constant rate \( \alpha \), the side length \( l \) at any time \( t \) can be expressed as:
\[ l = l_i - \alpha t \]
When the side length becomes \( a \), we can set:
\[ a = l_i - \alpha t \]
From this, we can solve for \( t \):
\[ t = \frac{l_i - a}{\alpha} \]

### Step 2: Calculate the area of the loop
The area \( A \) of the square loop at any time \( t \) is given by:
\[ A = l^2 = (l_i - \alpha t)^2 \]
Substituting \( t \) from the previous step, we have:
\[ A = (l_i - \alpha \cdot \frac{l_i - a}{\alpha})^2 = (a)^2 \]
Thus, the area of the loop when the side length is \( a \) is simply:
\[ A = a^2 \]

### Step 3: Determine the change in area over time
To find the induced EMF, we need to calculate the rate of change of area with respect to time:
\[ \frac{dA}{dt} = \frac{d}{dt}(l^2) = 2l \frac{dl}{dt} \]
Since \( \frac{dl}{dt} = -\alpha \) (the side is shrinking), we substitute \( l = a \):
\[ \frac{dA}{dt} = 2a(-\alpha) = -2a\alpha \]

### Step 4: Apply Faraday's law of electromagnetic induction
According to Faraday's law, the induced EMF \( \mathcal{E} \) in the loop is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux. The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A = B \cdot a^2 \]
Thus, the rate of change of magnetic flux is:
\[ \frac{d\Phi}{dt} = B \cdot \frac{dA}{dt} = B \cdot (-2a\alpha) \]

### Step 5: Calculate the induced EMF
Substituting this into Faraday's law gives:
\[ \mathcal{E} = -\left(-B \cdot 2a\alpha\right) = 2B a \alpha \]

Thus, the induced EMF in the loop when its side is \( a \) is:
\[ \mathcal{E} = 2B a \alpha \]

### Summary of the Solution
The induced EMF in the conducting square loop, when its side is \( a \), is given by:
\[ \mathcal{E} = 2B a \alpha \]