Let's call the +x direction to the left, the direction the bar is moving. The current through the bar is up. A current up in a field out of the page gives a force to the right:

Fx = ma = -ILB m dv/dt = -ILB The current is given by Ohm's Law, where the emf is the motional emf:

I = ε/R = vBL/R

Plugging this into our force expression gives:

m dv/dt = -vB2L2/R

The solution to this is the function that is basically the negative derivative of itself.

We need a negative exponential:

v(t) = vi e$^-t/τ

The time constant here is τ = mR/B^2.L^2

Let's call the $+x$ direction to the left, the direction the bar is moving. The current through the bar is up. A current up in a field out of the page gives a force to the right:

$$ F_x = ma = -ILB \Rightarrow\\ m \frac{dv}{dt} = -ILB. $$

The current is given by Ohm's Law, where the emf is the motional emf: $$ I = \frac{ε}{R} = \frac{vBL}{R} $$ Plugging this into our force expression gives: $$ m \frac{dv}{dt} = -\frac{vB^2L^2}{R} $$ The solution to this is the function that is basically the negative derivative of itself.

We need a negative exponential: $$ v(t) = v_i e^{-t/τ} $$ The time constant here is $τ = mR/(B^2L^2)$