The induced electromotive force for a coil having N turns in the coil and a change in flux \(\Delta \Phi \)in time \(\Delta t\) will be expressed as,
\(\begin{aligned}{}{\rm{\varepsilon }} &= - {\rm{N}}\frac{{{\rm{\Delta \Phi }}}}{{{\rm{\Delta t}}}}\\ &= - {\rm{N}}\frac{{{\rm{\Delta (BS)}}}}{{{\rm{\Delta t}}}}\end{aligned}\)……………..(1)
This we can discover by dialing the particular time t, as well as the fact that,
\({\rm{\Delta (BS)}} = {\rm{B\Delta S}} = {\rm{B}}\left( {{{\rm{S}}_{\rm{2}}}{\rm{ - }}{{\rm{S}}_{\rm{1}}}} \right) = {\rm{B}}\left( {{\rm{0 - }}{{\rm{S}}_{\rm{1}}}} \right) = {\rm{BS}}\)
Substitute the above values in equation (1) will give,
\(\varepsilon = {\rm{N}}\frac{{{\rm{BS}}}}{{\rm{t}}}\)……………….(2)
Now substitute the given values in the equation (2), we will get,
\(\begin{aligned}{}{\rm{\varepsilon }} &= {\rm{50 \times }}\frac{{\left( {{\rm{0}}{\rm{.250}}\;{{\rm{m}}^{\rm{2}}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.50}}\;{\rm{T}}} \right)}}{{{\rm{0}}{\rm{.100}}\;{\rm{s}}}}\\ &= {\rm{187}}{\rm{.5\;V}}\end{aligned}\)
The flux is decreasing as the field is directed into the page, implying that the induced current will have its field pointing into the page. Using the right-hand rule, we can determine that the current is flowing clockwise.
Therefore, the direction of the current is clockwise, and the value of EMF is \({\rm{187}}{\rm{.5\;V}}\)
