Question

The conducting rod $ab$, as shown in figure makes contact with metal rails $ca$ and $db$. The apparatus is in a uniform magnetic field of $0.800T$, perpendicular to the plane of the figure. Compare the rate at which mechanical work done by the force $\left(Fv\right)$ with the rate at which thermal energy is developed in the circuit ($I^{2}R$).(in $W$)

Answer 1

$F=F_m=ilB=\frac{e}{r}\cdot lB=\frac{Bvl}{r}\cdot Bl=\frac{B^2{l}^2}{r}\cdot v$$=\frac{{\left(0.8\right)}^2{\left(0.5\right)}^2}{{\left(1.5\right)}^{}}\times 7.5=0.8N$

$F\cdot v=0.8\times 7.5=6W$
$i^{2}R=\left(\frac{{\left(Bvl\right)}^2}{R}\right)=\frac{B^2{l}^2}{R}\cdot v^2$$=\frac{{\left(0.8\right)}^2{\left(0.5\right)}^2}{{\left(1.5\right)}^{}}\times {7.5}^2=6W$