Question

Two parallel vertical metallic rails $AB$ and $CD$ are separated by $1m$. They are connected at two ends by resistances $R_1$ and $R_2$ as shown in figure. A horizontal metallic bar of mass $0.2kg$ slides without friction vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of $0.6T$ perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in $R_1$ and $R_2$ are $0.76W$ and $1.2W$ respectively. Find the terminal velocity of the bar $L$ and the values of $R_1$ and $R_2$

Answer 1

Let the magnetic field be perpendicular to the plane of rails and inwards.
If $v$ be the terminal velocity of the rails, then potential difference across E and F would be $BvL$ with $E$ at lower potential and $F$ at higher potential.
The equivalent circuit is shown in figure (2). In figure (2)
$i_1=\frac{e}{R_1}$ ..(i)
$i_2=\frac{e}{R_2}$..(ii)
Power dissipated in $R_1$ is $0.76W$
Therefore $e{i}_1=0.76W$..(iii)
Similarly, $e{i}_2=1.2W$ ..(iv)
Now the total current in bar $EF$ is
$i=i_1+i_2$ ( from $E$ to $F$) ...( v)
or $i=3.27A$
Multiplying Eq. (v) by e, we get
$ei=e{i}_1+e{i}_2$
$=(0.76+1.2)$ [From Eqs. (iii) and (iv)]
$=1.96W$
$e=\frac{1.96}{i}V=\frac{1.96}{3.27}$
or $e=0.6V$
But since $e=BvL$
$v=\frac{e}{BL}=\frac{(0.6)}{(0.6)(1.0)}$
$=1.0m/s$.
Hence, terminal velocity of bar is $1.0m/s$.
Power in $R_1$ is $0.76W$
$\therefore 0.76=\frac{e^2}{R_1}$
$\Rightarrow \therefore R_1=\frac{e^2}{0.76}=\frac{(0.6{)}^2}{0.76}$
$=0.47\Omega \Rightarrow R_1=0.47\Omega$
Similarly, $R_1=\frac{1}{0.2}=\frac{(0.6{)}^2}{1.2}$
$=0.13\Omega$
$R_2=0.3\Omega$.