A circular loop made from a flexible, conducting wire is shrinking. Its radius as a function of time is $r=r_0{e}^{-\beta t}$. The loop is perpendicular to a steady, uniform magnetic field $B$. Find an expression for the induced emf in the loop at time $t$.

The electromagnetic effect on moving electric charges, electromagnetic fields, and magnetic materials is represented by a magnetic field, which is a vector field. In a magnetic flux, a moving charge generates a force that is perpendicular to both its own velocity and the magnetosphere.

From Faraday's law,

$\epsilon =\left|\frac{d{\Phi }_{\mathrm{m}}}{\epsilon }\right|.$

Where ${\Phi }_{\mathrm{m}}$is the flux through the loop which is the amount of magnetic field that flows through a loop of area localid="1648957060866" $A$and it is given by

${\Phi }_{\mathrm{m}}=BA$

Let us use this expression of localid="1648957068013" ${\Phi }_{\mathrm{m}}$into equation to get

$\epsilon =\frac{d\left(BA\right)}{dt}=B\frac{dA}{dt}.$

The radius of the loop $r$changes with time and we are given it in terms of time in the form

$r=r_o{e}^{-\beta t}$

The area of the circular loop is calculated by

$A=\pi r^2=\pi {\left(r_o{e}^{-\beta t}\right)}^2=\pi r_o^2{e}^{-2\beta t}$

Now, we use the expression of $A$into equation and differentiate this equation to get the emf by

$$\begin{gathered} \epsilon =B\frac{dA}{dt}=B\frac{d}{dt}\left(\pi r_o^2{e}^{-2\beta t}\right) \\ \epsilon ==B\pi r_o^2(-2\beta )e^{-2\beta t} \\ \epsilon =-2\pi r_o^2\beta B{e}^{-2\beta t}. \end{gathered}$$