Rotation + Magnetism = Rotatism

Let at any Time t=t rod is rotated by an angle ' $\theta$ ' So Consider an element of length 'dr' which is at an distance of 'r' from the point of Pivot ( Point P ).

Distance of This element from the wire is

$x\quad =\quad L+r\cos { \theta }$ .

Now emf induced between Point P and Centre O is :

${ \Ep }_{ P0 }\quad =\int _{ r=0 }^{ r=L/2 }{ \frac { { \mu }_{ 0 }{ I }_{ 0 }\omega \quad r }{ 2\pi (L+r\cos { \theta } ) } (dr) } \\ \\ { \Ep }_{ P0 }\quad =\quad \cfrac { { \mu }_{ 0 }{ I }_{ 0 }\omega }{ 2\pi } \int _{ r=0 }^{ r=L/2 }{ \frac { r }{ L+\cfrac { r }{ 2 } } (dr) } \quad \quad \quad \quad (\because \quad \cos { 60 } \quad =\quad 1/2\quad )\\ { \Ep }_{ P0 }=\quad \cfrac { { \mu }_{ 0 }{ I }_{ 0 }\omega }{ \pi } \quad \int _{ r=0 }^{ r=L/2 }{ \frac { 2L\quad +\quad r\quad -\quad 2L }{ 2L+r } (dr) } \\ \\ { \Ep }_{ P0 }=\quad \cfrac { { \mu }_{ 0 }{ I }_{ 0 }\omega }{ \pi } \quad \int _{ r=0 }^{ r=L/2 }{ (1\quad -\quad \frac { 2L }{ 2L+r } )(dr) } \\ \\ { \Ep }_{ P0 }=\quad \cfrac { { \mu }_{ 0 }{ I }_{ 0 }\omega }{ \pi } (\quad { [r\quad -\quad \frac { 2L }{ 2L+r } \ln { (2L+r) } ] }_{ 0 }^{ L/2 }\quad )\\ \\ { \Ep }_{ P0 }\quad =\frac { { \mu }_{ 0 }{ I }_{ 0 }\omega L }{ 4\pi \cos { \theta } } (1-\frac { 1 }{ \cos { \theta } } \ln { \frac { 25 }{ 16 } } )$ .

General solutions is given by $\frac { { \mu }_{ 0 }i\omega }{ 2\pi \cos { \theta } } \left( l-\frac { a }{ \cos { \theta } } \left( \ln { \left( \frac { a+l\cos { \theta } }{ a } \right) } \right) \right) \\ \\ Where\quad :\\ a\quad =\quad Distance\quad of\quad pivot\quad from\quad wire\\ l\quad =\quad length\quad of\quad rod\\ \theta \quad =\quad angle\quad made\quad by\quad rod\quad with\quad horizontal\\ \omega \quad =\quad Angular\quad frequency\\ i\quad =\quad Cirrent\quad in\quad wire$