Magnetism + Calculus + Mechanics = Megatulics

Well Now I feel very Tired Since It Takes approximately my 45 minutes to post it on brilliant. Anyway Solution Begins Here :

Diagram

Let any time t=t net induced emf in is ' ${ E }_{ ind }$ ' So

${ E }_{ ind }={ I }_{ ind }R$ .

Now Loop Is Super Conducting So $R\quad \approx \quad 0$ .

$\Rightarrow { \quad E }_{ ind }\quad =\quad 0$ .

So according To Faraday's Law of induction.

$\Rightarrow \quad \frac { -d{ \phi }_{ m } }{ dt } =\quad 0$ .

$\displaystyle{\therefore \quad { { \phi }_{ m } }_{ ,net,\quad at\quad t=t }\quad =\quad { { \phi }_{ m } }_{ ,net,\quad at\quad t=0 }\\ \\ \because { \quad \phi }_{ net }\quad =\quad { \phi }_{ external\quad mag.\quad field }\quad +{ \quad \phi }_{ self\quad current }\quad }$ .

Now Let at time t=t current in loop is ${ I }_{ ind }$ And Loop is at an height of y from the wire So Flux produced due to Self induction Phenomena will be :

$\displaystyle{{ { \quad (\phi }_{ self\quad current }) }_{ t=t }\quad =\quad L\times { I }_{ ind }\quad \longrightarrow (1)}$ .

Now For Calculating Flux from external magnetic field( mag. field of infinite wire) will be calculated by considering Rectangular strip ( green coloured shown in figure) of length ' $l$ ' and width dr which is at an height of ' r ' And at that instant of time Loop is at an height of y from wire :

$\displaystyle{{ { (\phi }_{ external\quad mag.\quad field }) }_{ t=t }=\quad \int _{ r=y }^{ r=y+l }{ { B }_{ ext } } .dA\cos { \pi } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\int _{ r=y }^{ r=y+l }{ { \frac { { \mu }_{ 0 }{ I }_{ 0 } }{ 2\pi r } .ldr } } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi } \ln { \frac { y+l }{ y } } \quad \longrightarrow (2)}$ .

Now net flux is constant So :

$\displaystyle{\because \quad { { \phi }_{ m } }_{ ,net,\quad at\quad t=t }\quad =\quad { { \phi }_{ m } }_{ ,net,\quad at\quad t=0 }\\ \Rightarrow \quad L{ I }_{ ind }\quad -\quad \frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi } \ln { \frac { y+l }{ y } \quad } =\quad -\frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi } \ln { \frac { 2l+l }{ 2l } } \\ \\ \Rightarrow \quad L{ I }_{ ind }\quad -\quad \frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi } \ln { \frac { y+l }{ y } \quad } =\quad -\frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi } \ln { \frac { 3 }{ 2 } } \\ \Rightarrow \quad { I }_{ ind }\quad =\quad \frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi L } \ln { \frac { 2(y+l) }{ 3y } \quad } \quad \longrightarrow (3)}$ .

Now Draw FBD of the loop at that instant of time. and use Newtons laws of motion

$\displaystyle{{ F }_{ net }\quad =\quad m\quad \times \quad { a }_{ net }}$ .

And You will see That net magnetic Force will acts in upwards and gravitational force in downward Direction.

$\displaystyle{mg\quad -{ \quad F }_{ mag }\quad =m(-\frac { vdv }{ dy } )\quad }$ .

Here I used the fact that

$\displaystyle{a\quad =\quad -\frac { vdv }{ dx } \quad \quad \quad \quad (\quad \because \quad y\downarrow es\quad with\quad v\uparrow es\quad )}$ .

So

$\displaystyle{mg\quad -\quad ({ I }_{ ind }\quad \times \quad l\quad \times \quad { B }_{ ext })\quad =-m\frac { vdv }{ dx } \quad \\ mg\quad -\quad ({ I }_{ ind }\quad \times \quad l\quad \times \quad \frac { { \mu }_{ 0 }{ I }_{ 0 } }{ 2\pi } (\frac { 1 }{ y } \quad -\quad \frac { 1 }{ y+l } )\quad =-m\frac { vdv }{ dx } \quad }$ .

Now Put all values which we have calculated already So the Differential equation becomes:

Note

here in solution I used frequently that

H=2 $l$

You check this according to data given in question.

$\displaystyle{{ (\frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi } })^{ 2 }\frac { 1 }{ mL } \int _{ y=2l }^{ y=l/2 }{ (\frac { 1 }{ y } } \quad -\quad \frac { 1 }{ y+l } )\ln { \frac { 2(y+l) }{ 3y } } dy\quad \quad -\quad g\int _{ y=2l }^{ y=l/2 }{ dy } \quad =\int _{ 0 }^{ v }{ vdv } \quad }$ .

Consider the integral $\displaystyle{E\quad =\quad \int _{ y=2l }^{ y=l/2 }{ (\frac { 1 }{ y } } \quad -\quad \frac { 1 }{ y+l } )\ln { \frac { 2(y+l) }{ 3y } } dy}$ .

Now Substitute the $\displaystyle{\ln { \frac { 2(y+l) }{ 3y } } \quad =\quad z}$ .

So this integral converts into :

$\displaystyle{(-\int { zdz } \quad =\quad -\frac { { z }^{ 2 } }{ 2 } \quad =\quad -\quad \frac { { (\ln { \frac { 2(y+l) }{ 3y } } ) }^{ 2 } }{ 2 } \\ \Rightarrow \quad E\quad =\quad -\quad { \left[ \frac { { (\ln { \frac { 2(y+l) }{ 3y } } ) }^{ 2 } }{ 2 } \right] }_{ 2l }^{ l/2 }\\ \\ \Rightarrow \quad \quad E=\quad \frac { -{ (\ln { 2 } ) }^{ 2 } }{ 2 } }$ .

So putting this integral in original equation we get an simply integrate The Rest part we get velocity

$\displaystyle{v\quad =\quad \sqrt { 3gl\quad -\quad { (\frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi } })^{ 2 }\frac { { (\ln { 2 } ) }^{ 2 } }{ mL } \quad } }$ .

Now put the Values we get answer i.e

$\\ v\quad =\quad 5.4 m\s$ .

Q.E.D

@Deepanshu Gupta your pain is nothing compared to mine , i spent 1 hour on it(solving+writing monstrous latex!), i read it wrong $$ , i thought you want time to be calculated , OH MAN ! when i read you want velocity , my face just brightened up and smile prevailed over it :P $$ now for the solution , $\frac { { d }^{ 2 }h }{ d{ t }^{ 2 } } =g-{ F }_{ mag. }$ here the current can be calculated as $d \phi$ = $di$ * $\L$ and then the force on current formula can be used , just to check you can see what ${ F }_{ mag. }$ comes , $$ it would be { I }_{ ind }\quad = \frac { { \mu }_{ 0 }{ I }_{ 0 }l }{ 2\pi L } \ln { \frac { 2(y+l) }{ 3y } \times \frac { { \mu }_{ 0 }{ I }_{ 0 } }{ 2\pi } (\frac { 1 }{ y } \quad -\quad \frac { 1 }{ y+l) } , $$ this can be equated to $m \cfrac {dv}{dx}$ , so , you were saved of that cumbersome double integral , you would have had to calculate ! :P thanks god !