A given wire is bent into a rectangular loop of size `15cm xx 5cm` and placed perpendicular to a magnetic field of 1.0 Tesla. Within 0.5sec, the loop is changed into a 10cm square and the field increases to 1.4 Tesla.Calculate the value of e.m.f. induced in the loop?

Answer

To calculate the induced electromotive force (e.m.f.) in the loop, we will follow these steps:

### Step 1: Calculate the initial area of the rectangular loop
The dimensions of the rectangular loop are given as 15 cm and 5 cm.

\[ \text{Area}_{\text{initial}} = \text{length} \times \text{width} = 15 \, \text{cm} \times 5 \, \text{cm} = 75 \, \text{cm}^2 \]

### Step 2: Convert the area from cm² to m²
To convert the area from cm² to m², we use the conversion factor \(1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2\).

\[ \text{Area}_{\text{initial}} = 75 \, \text{cm}^2 \times 10^{-4} \, \text{m}^2/\text{cm}^2 = 75 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Calculate the initial magnetic flux
The initial magnetic field \(B_i\) is given as 1.0 T. The magnetic flux \(\Phi_i\) is calculated using the formula:

\[ \Phi_i = B_i \times \text{Area}_{\text{initial}} = 1.0 \, \text{T} \times 75 \times 10^{-4} \, \text{m}^2 = 75 \times 10^{-4} \, \text{Wb} \]

### Step 4: Calculate the final area of the square loop
The dimensions of the square loop are given as 10 cm for each side.

\[ \text{Area}_{\text{final}} = \text{side} \times \text{side} = 10 \, \text{cm} \times 10 \, \text{cm} = 100 \, \text{cm}^2 \]

### Step 5: Convert the final area from cm² to m²
Again, we convert the area from cm² to m².

\[ \text{Area}_{\text{final}} = 100 \, \text{cm}^2 \times 10^{-4} \, \text{m}^2/\text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 \]

### Step 6: Calculate the final magnetic flux
The final magnetic field \(B_f\) is given as 1.4 T. The final magnetic flux \(\Phi_f\) is calculated as follows:

\[ \Phi_f = B_f \times \text{Area}_{\text{final}} = 1.4 \, \text{T} \times 100 \times 10^{-4} \, \text{m}^2 = 1.4 \times 10^{-2} \, \text{Wb} \]

### Step 7: Calculate the change in magnetic flux
The change in magnetic flux \(\Delta \Phi\) is given by:

\[ \Delta \Phi = \Phi_f - \Phi_i = (1.4 \times 10^{-2} \, \text{Wb}) - (75 \times 10^{-4} \, \text{Wb}) \]

Converting \(75 \times 10^{-4}\) Wb to the same unit as \(\Phi_f\):

\[ 75 \times 10^{-4} \, \text{Wb} = 0.0075 \, \text{Wb} \]

Thus,

\[ \Delta \Phi = 0.014 \, \text{Wb} - 0.0075 \, \text{Wb} = 0.0065 \, \text{Wb} \]

### Step 8: Calculate the induced e.m.f.
The induced e.m.f. (\(E\)) can be calculated using Faraday's law of electromagnetic induction:

\[ E = -\frac{\Delta \Phi}{\Delta t} \]

Given that \(\Delta t = 0.5 \, \text{s}\):

\[ E = -\frac{0.0065 \, \text{Wb}}{0.5 \, \text{s}} = -0.013 \, \text{V} \]

Since we are interested in the magnitude of e.m.f., we can ignore the negative sign:

\[ E = 0.013 \, \text{V} \]

### Final Answer
The value of the induced e.m.f. in the loop is **0.013 V**.

---