Two perfectly conducting rails are placed on the $x y$-plane, one coincident with the $x$-axis, starting at $(0,0)$, the other parallel to the first rail a distance $\ell$ apart, starting at $(0, \ell)$. A resistor $R$ is connected across the rails between $(0,0)$ and $(0, \ell)$, and a uniform magnetic field $\mathbf{B}=B \hat{\mathbf{e}}_{z}$, where $\hat{\mathbf{e}}_{z}$ is the unit vector along the $z$-axis and $B>0$, fills the entire region of space. A metal bar of negligible resistance and mass $m$ slides without friction on the two rails, lying perpendicular to both of them in such a way that it closes the circuit formed by the rails and the resistor. The bar moves with speed $v$ to the right such that the area of the loop becomes larger with time.

(i) Calculate the current in the resistor and indicate its direction of flow in a diagram of the system.

(ii) Show that the magnetic force on the bar is

$$\mathbf{F}=-\frac{B^{2} \ell^{2} v}{R} \hat{\mathbf{e}}_{x}$$

(iii) Assume that the bar starts moving with initial speed $v_{0}$ at time $t=0$, and is then left to slide freely. Using your result from part (ii) and Newton's laws show that its velocity at the time $t$ is

$$v(t)=v_{0} e^{-\left(B^{2} \ell^{2} / m R\right) t} .$$

(iv) By calculating the total energy delivered to the resistor, verify that energy is conserved.