The figure shows a square loop of wire with sides of length $l=2cm$. A magnetic field points  into the page and its magnitude is given by $B=a{t}^{2}y$ where $a=4T{m}^{-1}{s}^{-2},$B is in tesla, t is in  second and y is in meter. Determine the emf induced in the loop at t =2.5sec is ……… $\mu V$

 

Here we note that the field is a simultaneous function of y and t. Let us first calculate the  flux associated with the loop. For that let us consider a rectangular strip of length l,  thickness by at a distance y from the x-axis. Then the flux associated with strip is $d\varphi =BdA=Bldy\Rightarrow d\varphi =a{t}^{2}yldy$

$\Rightarrow \varphi =\int d\varphi =a{t}^2\underset{0}{\overset{1}{\int }}ydy\Rightarrow \varphi =\frac{a{t}^2{l}^3}{2} \{\because \underset{0}{\overset{1}{\int }}ydy=\frac{l^2}{2}\}$

Now, according to Faraday’s Laws, we have $\left|e\right|=\frac{d\varphi }{dt}$

At t=2.5 sec

emf=80 $\mu V$