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(B) The circuit can be modeled as an inductor $L$ in series with a motional electromotive force $(EMF)$ $e = B l v$,where $l$ is the length of the arm

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(B) The circuit can be modeled as an inductor $L$ in series with a motional electromotive force $(EMF)$ $e = B l v$,where $l$ is the length of the arm in the magnetic field. Since the loop is rectangular and moving with constant velocity $v$,the length of the arm in the field is constant,let it be $l$. Applying Kirchhoff's voltage law to the loop:$e - L \frac{dI}{dt} = 0$$v B l - L \frac{dI}{dt} = 0$$\frac{dI}{dt} = \frac{v B l}{L}$ ..... $(i)$Given that the distance of the left arm from the origin is $x = vt$,we have:$\frac{dx}{dt} = v$ ..... $(ii)$Using the chain rule,$\frac{dI}{dx} = \frac{dI}{dt} \cdot \frac{dt}{dx} = \frac{dI}{dt} \cdot \frac{1}{v}$.Substituting from $(i)$:$\frac{dI}{dx} = \left( \frac{v B l}{L} \right) \cdot \frac{1}{v} = \frac{B l}{L}$.Since $B, l,$ and $L$ are constants,the slope $\frac{dI}{dx}$ is a positive constant. Therefore,the current $I$ increases linearly with distance $x$. The correct graph is a straight line passing through the origin (assuming $I=0$ at $x=0$).

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