$[$The "perfect" inductive response$]$ Consider the arrangement shown in the figure below, where there is a uniform magnetic field perpendicular to the plane of the figure, only in the hatched region. Suppose the rectangular loop, with sides $h$ and $w,$ is moving at a constant velocity $v,$ as indicated. We saw in class that the total electromotive force $($emf$)$ in the circuit of the loop can be understood as the sum of two terms:

$E=E_{\text{f}\text{l}\text{u}\text{x}}+E_{\text{s}\text{e}\text{l}\text{f}-\text{i}\text{n}\text{d}\text{.}}=-\frac{d{}_B}{dt}-L\frac{dI}{dt},$

where the first term corresponds to the flux variation due to the external field, and the second term corresponds to the emf induced by the self$-$inductance of the circuit itself $(L$ is the selfinductance of the loop$).$ If the wire has an internal resistance $R,$ we can set the expression above equal to $RI$ to obtain a differential equation for the function $I(t).$

However, if the wire's resistance can be neglected, the total emf can be set to zero, meaning that $E_{\text{s}\text{e}\text{l}\text{f}-\text{i}\text{n}\text{d}\text{.}}=-E_{\text{f}\text{l}\text{u}\text{x}}.$ In other words, neglecting the wire's resistance represents an idealized situation where the self$-$inductance response exactly cancels out the emf induced by the flux variation.

In this case, explicitly calculate $E_{\text{f}\text{l}\text{u}\text{x}}$ in terms of $B,v,h,$ and $t$ as a function of time $t,$ and find the current in the loop $I(t)$ for $t>0.$ Assume the loop starts leaving the hatched region at $t=0$ with $I(0)=0.$ Plot $I(t).$ Finally, calculate the magnetic force on the loop due to the external field as a function of time.