A conducting square frame of side $$'a'$$ and a long straight wire carrying current $$I$$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity $$'v'.$$ The emf induced in the frame will be proportional to

Solution :

Potential difference across $$PQ$$  is
$${V_P} - {V_Q} = {B_1}\left( a \right)v = \frac{{{\mu _0}I}}{{2\pi \left( {x - \frac{a}{2}} \right)}}av$$
Potential difference across side $$RS$$  of frame is
$${V_S} - {V_R} = {B_2}\left( a \right)v = \frac{{{\mu _0}I}}{{2\pi \left( {x + \frac{a}{2}} \right)}}av$$
Hence, the net potential difference in the loop will be
$${V_{{\text{net}}}} = \left( {{V_P} - {V_Q}} \right) - \left( {{V_S} - {V_R}} \right)$$
$$\eqalign{ & = \frac{{{\mu _0}iav}}{{2\pi }}\left[ {\frac{1}{{\left( {x - \frac{a}{2}} \right)}} - \frac{1}{{\left( {x + \frac{a}{2}} \right)}}} \right] \cr & = \frac{{{\mu _0}iav}}{{2\pi }}\left( {\frac{a}{{\left( {x - \frac{a}{2}} \right)\left( {x + \frac{a}{2}} \right)}}} \right) \cr} $$
Thus, $${V_{{\text{net}}}} \propto \frac{1}{{\left( {2x - a} \right)\left( {2x + a} \right)}}$$