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Area of loop $$AEBCFDA =$$ area of rectangle $$ABCD \ -$$ Area of triangle $$ABE\ -$$ Area of triangle $$DCF$$

$$\therefore$$ Required area $$=16cm \times 4 cm - \dfrac{1}{2}\times 4cm \times 2cm - \dfrac{1}{2}\times 4cm \times 2cm$$

$$= 64cm^2 - 4cm^2 - 4cm^2 = 56cm^2$$

Magnetic flux passing through an area is given by

$$\displaystyle \phi = \int B.dA = \int BdA \cos \theta$$.

Here, $$\theta = 0$$ and $$B$$ is constant throughout the area.

$$\therefore \phi_{initial} = B_{initial}.A = 1000 \, Gauss \times 56cm^2$$

$$= 56000 \, gauss\, cm^2$$

$$= 5.6\times 10^{-4}Wb$$

$$\begin{bmatrix} 1 \, Gauss = 10^{-4}T \, and \, 1cm^2 = 10^{-4}m^2\\ \therefore 1\, Gauss\, cm^2 = 10^{-8}Tm^2 = 10^{-8}Wb \end{bmatrix}$$

Also, $$\phi _{final} = B_{final}. A = 500\, Gauss \times 56cm^2$$

$$= 28000\, Gauss\, cm^2 = 2.8\times 10^{-4}Wb$$

Since the flux changes linearly,

$$\in = \left|\dfrac{d\phi}{dt}\right| = \dfrac{|\phi_{final} - \phi_{initial}|}{\Delta t} = \dfrac{(5.6-2.8)\times 10^{-4}}{5}V$$

$$= 56\times 10^{-6}V = 56\mu v$$

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