Question

A wire 40 cm long is bent into a rectangular frame 15 cm $\times 5cm$ and placed perpendicular to a field of induction 0.8 $T$ in 0.5 second. The frame is change into a square frame and the filed is increased to 1.4 T. Then e.m.f induced in the frame is

• A. $1.6\times {10}^{-2}V$
• B. $1.6\times {10}^{3}V$
• C. $1.6\times {10}^{-4}V$
• D. $1.6\times {10}^{-5}V$

Answer 1

The correct option is A $1.6\times {10}^{-2}V$

Given that,

Field of induction B₁ = 0.8T

Area A₁ = (15×5×10^-4) m²

Now, the flux

${\varphi }_1=75\times {10}^{-4}\times 0.8$

${\varphi }_1=60\times {10}^{-4}Wb$

Increases field B₂ = 1.4T

Area of square A₂ = (10×10×10^-4) m²

For square,

Flux

${\varphi }_2=B_2{A}_2$

${\varphi }_2=(1.4\times 100\times {10}^{-4})Wb$

Change in magnetic flux

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

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

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

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

Now, Induced emf is

$e=\frac{\Delta \varphi }{\Delta t}$

$e=\frac{80\times {10}^{-4}}{0.5}$

$e=160\times {10}^{-4}$

$e=1.6\times {10}^{-2},volt$

Hence, the e.m.f induced in the frame is $e=1.6\times {10}^{-2}$ volt.