Why are you in a hurry?

In this problem , our primary aim is to find the steady state velocity. This can be done by two methods, method one is more important, as it involves more learning.

_ Method-1 : Finding v(t) _

Let us consider the given figure. Clearly the current will start flowing anticlockwise. Hence, lorentz force would act forward. Also, there will be an induced EMF due to the motion of rod. Let, at a time $t$ , the current is $i$ , charge at capacitor is $q$ , and the velocity of rod is $v$ .

Clearly, $i = \frac{-dq}{dt}$ (as $q$ is decreasing.).

Also, the lorentz force is $iLB$ , hence, the rate of change of velocity,

$\frac{dv}{dt} = \frac{iLB}{M} = -\frac{dq}{dt} \frac{BL}{M}$ , note that $i = \frac{M}{BL} \frac{dv}{dt}$

Hence, we get : $dv = -\frac{BL}{M} dq$ ,

Integrate to get:

$v = \frac{BL}{M}(Q-q) \Rightarrow q = Q - \frac{Mv}{BL}$

Now, we apply kirchoff's voltage law in the circuit, also considering the induced EMF, which clearly equals $BvL$ .

$BvL + iR - \frac{q}{C} = 0$

$\displaystyle \Rightarrow BvL + \frac{MR}{BL} \frac{dv}{dt} - \frac{(Q - \frac{Mv}{BL})}{C} = 0$

Rearrange to get:

$\displaystyle \frac{dv}{BLQ - (B^2L^2C +M)v} = \frac{dt}{MRC}$

$\Rightarrow \displaystyle \int_{0}^{v} \frac{dv}{BLQ - (B^2L^2C +M)v} = \int_{0}^{t} \frac{dt}{MRC}$

$\Rightarrow$ $\boxed{\displaystyle v(t) = \frac{BLQ}{B^2L^2C + M} \bigg(1 - e^{- \frac{B^2L^2C+M}{MRC} t} \bigg)}$

Clearly, $v$ in steady state would be $v_{0} = \frac{BLQ}{B^2L^2C + M}$ .

Method-2 : Effortless.

Clearly, in steady state, the velocity becomes constant and hence the net force becomes $0$ . Thus, $i$ also becomes $0$ . Hence, using Kirchoff's voltage law equation, we get:

$q = Bv_{0} LC$ .

But we have already proved that $v = \frac{BL}{M} (Q - q)$

$\Rightarrow v_{0} = \frac{BL}{M} (Q- Bv_{0} LC)$ .

Solve to get $\boxed{v_{0} = \frac{BLQ}{B^2L^2C + M}}$

Maximising kinetic energy

Clearly, kinetic energy at steady state is $K = \frac{1}{2} Mv_{0}^2 = \frac{B^2L^2Q^2}{2} \frac{M}{(B^2L^2C+M)^2}$

To maximise $K$ , we have to maximise $\frac{M}{(B^2L^2C+M)^2}$ , as other term is constant.

Rest part involves taking derivative with respect to $M$ and putting $0$ , i.e.

$\frac{M - B^2L^2C}{(M+B^2L^2C)^2} = 0$ $\Rightarrow \boxed{M = B^2L^2C}$ .

Put values to get $M = 4$ mg

Let, at a general time $t$ ,
Current through the rod $= i$

Then,
we know that the Force on the Rod $= BiL$

Hence,
$BiL = Ma$ , where $a$ is the acceleration of the Rod
$\Rightarrow BL\frac{dq}{dt} = M\frac{dv}{dt}$

$\Rightarrow BL\int_Q^q dq = M\int_0^v dv$
$\Rightarrow BL(q-Q) = Mv$

Also,
When the velocity of the Rod is $v$ , the EMF induced in the rod is $BLv$
Hence,
$q=CV=C(BLv)$

Substituting the value of $q$ ,
$BL(CBLv-Q)=Mv$

$\Rightarrow v=\frac{BLQ}{M+CB^2L^2}$

Thus,
Kinetic Energy K $=\frac{1}{2}M(\frac{BLQ}{M+CB^2L^2})^2$

Differentiating with respect to $M$ ,
$\frac{dK}{dM} = 0$

$\frac{1}{2}-\frac{M}{M+CB^2L^2}=0$

$\Rightarrow M=4\times10^{-6}Kg = \boxed{4mg}$