Also, \(\varepsilon=\mathrm{i} \mathrm{R}\)
\(\therefore \quad i R=\frac{d \phi}{d t} \quad \Rightarrow \int d \phi=R \int i d t\)
Magnitude of change in flux \((\mathrm{d} \phi)=R \times\) area under current vs time graph
or, \(\quad d \phi=100 \times \frac{1}{2} \times \frac{1}{2} \times 10=250\, \mathrm{Wb}\)