A wire 40cm long is bent into a rectangular frame 15cm × 5 cm and placed perpendicular to a field of induction 0.8T. In 0.5 sec, the frame is changed into a square frame and the field is increased to 1.4T. Then e.m.f induced in the frame is 
 

Given data:

Field of induction ${\mathrm{B}}_1=0.8 \mathrm{T}$

Increases field ${\mathrm{B}}_2=1.4 \mathrm{T}$

Concept used: Flux and induced EMF

Detailed solution:

Field of induction ${\mathrm{B}}_1=0.8 \mathrm{T}$

Area ${\mathrm{A}}_1=\left(15\times 5\times {10}^{-4}\right){\mathrm{m}}^2$

Now, the flux for rectangular frame is

$$\begin{gathered} {\varphi }_1=75\times {10}^{-4}\times 0.8 \\ {\varphi }_1=60\times {10}^{-4} \text{Wb} \end{gathered}$$

Increases field ${\mathrm{B}}_2=1.4 \mathrm{T}$

Area of square ${\mathrm{A}}_2=\left(10\times 10\times {10}^{-4}\right){\mathrm{m}}^2$

Now the Flux for square frame is 

${\varphi }_2={\text{B}}_2 {\text{A}}_2\text{; }{\varphi }_2$ $=\left(1.4\times 100\times {10}^{-4}\right)\text{Wb}$

Change in magnetic flux

$\Delta \varphi ={\varphi }_2-{\varphi }_1$

$\Delta \varphi \text{ =}1.4\times 100\times {10}^{-4}-0.8\times 75\times {10}^{-4}$

$\Delta \varphi =(140-60)\times {10}^{-4}$

$\Delta \varphi =80\times {10}^{-4} \text{Wb}$

Now, Induced EMF is

$$\begin{gathered} \text{e}=\frac{\Delta \varphi }{\Delta t} \\ \text{e=}\frac{80\times {10}^{-4}}{0.5}=1.6\times {10}^{-2}\mathrm{volt} \end{gathered}$$