Question

Two parallel vertical metallic rails $AB$ and $CD$ are separated by $1\text{m}.$ They are connected at the two ends by resistances $R_1$ and $R_2$ as shown in figure. A horizontal metallic bar $PQ$ of mass $0.2\text{kg}$ slides without friction with maintaining contacts with the rails, vertically downward under the action of gravity. There is a uniform horizontal magnetic field of $0.6\text{T}$ perpendicular to the plane of the rails in the region. It is observed when the terminal velocity is attained, the powers dissipated in $R_1$ and $R_2$ are $0.76\text{W}$ and $1.2\text{W}$ respectively.
The terminal velocity of the bar is $n\text{m/s}$. The value of $n$ is (Take $g=9.8{\text{m/s}}^2)$

Answer 1

Due to sliding of the bar, an EMF is induced which causes currents to flow in the resistances as shown in the figure below.


The bar will acquire terminal velocity $V_T$, when:
$mg=Bil$

$\Rightarrow i=\frac{mg}{Bl}=\frac{0.2\times 9.8}{0.6\times 1}=\frac{49}{15}\text{A}$

Here $i=\frac{49}{15}=i_1+i_2----\left(1\right)$

Also, induced emf, $E=i_1{R}_1=i_2{R}_2------\left(2\right)$

It is given that,
Power dissipated in resistance $R_1$ is $P_1=0.76\text{W}$

$\Rightarrow P_1=E{i}_1=0.76-----\left(3\right)$

and Power dissipated in resistance $R_2$ is
$P_2=E{i}_2=1.2-------\left(4\right)$

$\therefore \frac{P_1}{P_2}=\frac{i_1}{i_2}=\frac{0.76}{1.2}$

$\Rightarrow \frac{i_1}{i_2}=\frac{19}{30}$

$\Rightarrow i_1=\frac{19}{30}{i}_2--------\left(5\right)$

From $\left(1\right)$ and $\left(5\right)$,

$i_1=\frac{19}{15}\text{A}$ and $i_2=2\text{A}$

from eqn $\left(4\right)$
$E=\frac{1.2}{2}=0.6\text{V}$

$\therefore$ Terminal speed is given by

$V_T=\frac{E}{Bl}=\frac{0.6}{0.6\times 1}=1\text{m/s}$

Hence, $n=1$