A conducting wire $20 c m$ long, bent into rectangular loop $6 c m \times 4 c m$ is placed(kept) perpendicular to the magnitude field of induction $0.5 T$ . Within $0.1 s$ . the loop is changed to a square and magnitude induction is increases to $1 T$ . What will be the emf induced in the wire?
Data Given:
$A_{1} = 24 c m^{2} = 24 \times 10^{- 4} m^{2}$
$A_{2} = 25 \times 10^{- 4} m^{2}$
$B_{1} = 0.5 T$
$B_{2} = 1 T$
$d t = 0.1 s$

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Solution

Emf Induced $= \frac{△ ϕ_{B}}{△ T}$
$△ ϕ_{B} =$ Change in Flux.
$ϕ_{f} =$ Final flux $= (1) (25 \times 10^{- 4})$
$ϕ_{i} =$ Initial Flux $= (0.5) (24 \times 10^{- 4})$
$△ ϕ_{B} - ϕ_{f} - ϕ_{i} = 25 \times 10^{- 4} - 12 \times 10^{- 4}$
$= 13 \times 10^{- 4}$
Emf $= \frac{13 \times 10^{- 4}}{0.1} = 13 \times 10^{- 3} V o l t s$

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Emf Induced=△ϕB△T△ϕB= Change in Flux.ϕf= Final flux=(1)(25×10−4)ϕi= Initial Flux =(0.5)(24×10−4)△ϕB−ϕf−ϕi=25×10−4−12×10−4=13×10−4Emf=13×10−40.1=13×10−3Volts