Question

Two parallel conducting rails are connected to a source of emf $E$ and internal resistance $r.$ Another conducting rod of length $l$ having negligible resistance lies at rest and can slide without friction over the rails. A uniform magnetic field $B$ is applied perpendicular to the plane of the rails. At $t=0,$ the rod is pulled along the rails by applying a force $F.$ The velocity of the rod is observed to be $v=v_{0}cos\left(\omega t\right)$ then find the average power (in watt) spent by the force over $1$ cycle. (Given $B=2$ Tesla, $r=2\times 10^{-4}\Omega ,v_0=2m{s}^{-1},l=2cm$ )

Answer 1

Answer: 4

$Blv-E-ir=0$
$F-ilB=m\frac{dv}{dt}\Rightarrow i=\frac{Blv-E}{r}$
$P_{avg}=\frac{\int \overrightarrow{F}\cdot \overrightarrow{v}dt}{\int dt}$
$P_{avg}=\frac{\int \left(m\frac{dv}{dt}+ilB\right)vdt}{\int dt}=\frac{\int \frac{l^2{B}^2}{r}{v}^{2}dt}{\int dt}$
$=\frac{v^2{l}^2{B}^2}{2r}=\frac{4\times 4\times 10^{-4}}{2\times 2\times 10^{-4}}=4$