Faraday's Law Of Induction

According to NCERT Class 12, Chapter 6 (Electromagnetic Induction), Faraday's law states that the induced emf equals the negative rate of change of magnetic flux linkage, expressed as $\epsilon =-N\frac{d\Phi }{dt}$, where flux $\Phi =BA$ for a field perpendicular to the coil. The area of the coil is $A=\pi r^2=\pi (0.05{)}^2=7.854\times 10^{-3}{\text{m}}^2$. The change in flux per turn is $\Delta \Phi =A\Delta B=7.854\times 10^{-3}\times (0.6-0.2)=3.142\times 10^{-3}\text{Wb}$. The induced emf is $\epsilon =N\frac{\Delta \Phi }{\Delta t}=200\times \frac{3.142\times 10^{-3}}{0.4}=1.571\text{V}$. The option 0.785 V wrongly omits the factor of 2 from the field change or halves the turns. The option 0.393 V ignores the number of turns entirely. The option 3.142 V forgets to divide by the time interval. A quick sanity check confirms units: (turns × T × m² / s) yields volts, and the magnitude of about 1.6 V is reasonable for 200 turns and a rapid field change. It is worth emphasising that only the change in flux matters, not the flux itself, so a steady field of any strength would induce no emf at all. Here the field falls at a uniform rate, so the induced emf stays constant throughout the interval rather than varying with time. The negative sign in Faraday's law, captured by Lenz's law, tells us the induced current would flow so as to oppose the decrease in flux, effectively trying to sustain the original field.