Problem 23.45

23.45
The current in the bar is given by
$I = \frac{\epsilon}{R} + \frac{\epsilon_{induced}}{R}$
where
$\epsilon_{induced} = -\frac{d}{dt} (BA)$
The force due to the emf is given by
F_emf = BId
to find the acceleration of the rod,
BId = ma
Solving for the acceleration, we find that
$a = \frac{IBd}{m}$
$a = \frac{Bd}{mR} (\epsilon + \epsilon_{induced})$
$a = \frac{Bd}{mR} (\epsilon - Bvd)$
But acceleration can be written as
$a = \frac{dv}{dt}$
Therefore, we have a first order differential equation.
$\frac{dv}{dt} = \frac{Bd}{mR} (\epsilon - Bvd)$
To solve this, make a change of variables. Let
$u = \epsilon - Bvd$
Then taking a derivative, we have that
$\frac{du}{dt} = -Bd \frac{dv}{dt}$
$\frac{1}{Bd} \frac{du}{dt} = \frac{Bd}{mR} u$
$\int_{u_{0}}^{u} \frac{du}{u} = -\int_{t=0}^{t}\frac{(Bd)^{2}}{mR} dt$
Integrating, we find that
$\ln \frac{u}{u_{0}} = - \frac{(Bd)^{2}}{mR} t$
u = u₀ e^{-B<<168>>2d<<169>>2t/mR}
Since v = 0 when t = 0, then
$u_{0} = \epsilon$
$u = \epsilon - Bvd$
Therefore,
$\epsilon - Bvd = \epsilon e^{-B^{2}d^{2}t/mR}$
$v = \frac{\epsilon}{Bd}(1 - e^{-B^{2}d^{2}t/mR})$