A conducting frame $abcd$ is kept in a vertical plane. A conducting rod $ef$ of mass $m$ and length $l$ can slide smoothly on it remaining always horizontal. The resistance of the loop is negligible and inductance is constant having value $L$. The rod is left from rest and allowed to fall under gravity and inductor has no initial current. A magnetic field of constant magnitude $B$ is present throughout the loop pointing inwards. Determine

 

(a) position of the rod as a function of time assuming initial position of the rod to be $x=0$ and vertically downward as the positive x-axis. 

(b) the maximum current in the circuit. 

(c) maximum velocity of the rod

Suppose $v$ be the velocity of the rod $ef$ when it has fallen a distance $x$. Then, 

$V_{fe}=V_{cb}$ or $Bvl=L\left(di/dt\right)$

or $B\left(dx/dt\right)l=L\left(di/dt\right)$ or $Bl\left(dx\right)=L\left(dt\right)$

Integrating, we get $Li=Blx$

or $i=\left(\frac{Bl}{L}\right)x$ ………… (i)

Now, magnetic force opposite to displacement $x$ will be

$F=F_m=ilB=\left(\frac{B^2{l}^2}{L}\right)x$

A constant downward force is $mg$. 

So, this is similar situation like spring-block system in vertical position. In which a force $F=kx$ acts upwards and a constant force $mg$ acts downwards. 

Hence, the wire will execute SHM, where

$k=\frac{B^2{l}^2}{L}$

Amplitude will be at $F_m=mg$

 or $\left(\frac{B^2{l}^2}{L}\right)A=mg \Rightarrow A=\frac{mgL}{B^2{l}^2}$

At $t=0$, rod is in its extreme position. 

Therefore, if we write the equation from mean position we will write, 

$X=-A\cos \omega t$

But, $x=X+A=A-A\cos \omega t=A\left(1-\cos \omega t\right)$

where, $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{B^2{l}^2}{mL}}$

(b) From Eq.(i), 

$i_{max}=\left(\frac{Bl}{L}\right)x_{max}$

Here, $x_{max}=2A=\frac{2mgL}{B^2{l}^2}$

$\therefore i_{max}=\left(\frac{Bl}{L}\right)\left(\frac{2mgL}{B^2{l}^2}\right)=\frac{2mg}{Bl}$

(c) Maximum velocity, 

$v_0=\omega A=\left(\sqrt{\frac{B^2{l}^2}{mL}}\right)\left(\frac{mgL}{B^2{l}^2}\right)=g\sqrt{\frac{mL}{B^2{l}^2}}=\frac{g\sqrt{mL}}{Bl}$