Simple isn't it? #11

$Variable\quad of\quad magnetic\quad field\quad used\quad here=B\\ Variable\quad of\quad electric\quad field\quad used\quad here=E\\ Variable\quad of\quad magnetic\quad flux\quad used\quad here=\phi \\ Variable\quad of\quad no\quad of\quad windings\quad used\quad here=B\\ For\quad convinence\quad we\quad take\quad I=kt\quad and\quad i=\alpha kt,\quad k=20A{ s }^{ -1 }\quad \alpha =2\\ { B }_{ I }={ \mu }_{ 0 }{ n }_{ I }{ I }\\ { B }_{ i }={ \mu }_{ 0 }{ n }_{ i }{ i }\\ \therefore { \phi }_{ I }=\quad { \mu }_{ 0 }{ n }_{ I }{ I }.\pi { R }^{ 2 },\quad \left[ R\quad is\quad what\quad we\quad need\quad to\quad find \right] \\ \therefore { \phi }_{ i }=\quad { \mu }_{ 0 }{ n }_{ i }{ i }.\pi { a }^{ 2 }\\ We\quad know\quad that,\\ \oint { E.dl\quad = } \frac { -d\phi }{ dt } \\ \therefore \quad E\quad =\quad \frac { { \mu }_{ 0 }k\left[ { n }_{ I }{ R }^{ 2 }-\alpha { n }_{ i }{ a }^{ 2 } \right] }{ 2R } \\ Since\quad E\quad is\quad constant\quad eE\quad is\quad also\quad constant,\\ \therefore \quad On\quad balancing\quad all\quad the\quad forces,\\ we\quad get\quad R=a\sqrt { \frac { { n }_{ i }\alpha }{ { n }_{ I } } } \\ \therefore R=2m\quad \\ At\quad least\quad balancing\quad of\quad forces\quad should\quad be\quad done\quad by\quad yourself.$

Easy. What we want is that e must keep revolving at that R. Therefore inducedEMF =0 So we will equate d(flux from an area pi.r²)/dt for solenoid 1 and solenoid 2 so that there will be constant B and hence e will revolve at same R