A conducting wire is bent in the form of a square of side 8 cm and is placed in a region of magnetic field of 0.05 T, such that the plane of coil is perpendicular to the magnetic field. If the side of square starts decreasing at a rate of `10^(-3) m s^(-1)`, what will be the emf induced in the coil when the area of coil is reduced to just half?

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the problem
We have a square coil made of a conducting wire with an initial side length of 8 cm. The coil is placed in a magnetic field of 0.05 T, and the side of the square is decreasing at a rate of \(10^{-3} \, \text{m/s}\). We need to find the induced EMF when the area of the coil is reduced to half its original value.

### Step 2: Convert units
Convert the side length of the square from centimeters to meters:
\[ \text{Side length} = 8 \, \text{cm} = 8 \times 10^{-2} \, \text{m} \]

### Step 3: Calculate the initial area of the square
The area \(A\) of the square is given by:
\[ A = \text{side}^2 = (8 \times 10^{-2})^2 = 64 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Determine the area when it is halved
When the area is reduced to half, the new area \(A'\) is:
\[ A' = \frac{A}{2} = \frac{64 \times 10^{-4}}{2} = 32 \times 10^{-4} \, \text{m}^2 \]

### Step 5: Find the new side length corresponding to the halved area
Let the new side length be \(a'\). Then:
\[ a'^2 = 32 \times 10^{-4} \implies a' = \sqrt{32 \times 10^{-4}} = \sqrt{32} \times 10^{-2} \, \text{m} \]

### Step 6: Calculate the magnetic flux
The magnetic flux \(\Phi\) through the coil is given by:
\[ \Phi = B \cdot A \]
Where \(B\) is the magnetic field strength. Thus, the initial flux is:
\[ \Phi = 0.05 \, \text{T} \cdot (64 \times 10^{-4} \, \text{m}^2) = 0.05 \cdot 64 \times 10^{-4} = 3.2 \times 10^{-5} \, \text{Wb} \]

### Step 7: Determine the rate of change of area
The side length is decreasing at a rate of \(10^{-3} \, \text{m/s}\). Therefore, the rate of change of area \( \frac{dA}{dt} \) can be calculated as:
\[ \frac{dA}{dt} = 2a \cdot \frac{da}{dt} \]
Substituting \(a = 8 \times 10^{-2} \, \text{m}\) and \(\frac{da}{dt} = -10^{-3} \, \text{m/s}\):
\[ \frac{dA}{dt} = 2(8 \times 10^{-2}) \cdot (-10^{-3}) = -1.6 \times 10^{-4} \, \text{m}^2/\text{s} \]

### Step 8: Calculate the induced EMF
The induced EMF (\(\mathcal{E}\)) is given by Faraday's law:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -B \cdot \frac{dA}{dt} \]
Substituting the values:
\[ \mathcal{E} = -0.05 \cdot (-1.6 \times 10^{-4}) = 8 \times 10^{-6} \, \text{V} \]

### Final Answer
The induced EMF when the area of the coil is reduced to half is:
\[ \mathcal{E} = 8 \times 10^{-6} \, \text{V} \text{ or } 8 \, \mu\text{V} \]