Derive the equation for the induced $emf$ in a rod of length $l$ sliding with velocity $v$ on a $U$-shaped frame placed perpendicular to a uniform magnetic field $B$.

(N/A) Consider a conducting rod $PQ$ of length $l$ moving with a constant velocity $v$ to the left on a $U$-shaped conducting frame.
Let the magnetic field $B$ be uniform and directed perpendicular to the plane of the frame into the page.
The area $A$ of the loop formed by the rod and the frame is $A = l \times x$,where $x$ is the distance of the rod from the left end of the frame.
The magnetic flux $\Phi_B$ linked with the loop is given by $\Phi_B = B \cdot A = B \cdot l \cdot x$.
According to Faraday's law of electromagnetic induction,the induced $emf$ $\varepsilon$ is given by $\varepsilon = -\frac{d\Phi_B}{dt}$.
Substituting the expression for $\Phi_B$,we get $\varepsilon = -\frac{d}{dt}(B \cdot l \cdot x)$.
Since $B$ and $l$ are constant,$\varepsilon = -B \cdot l \cdot \frac{dx}{dt}$.
As the rod moves to the left,the distance $x$ decreases,so $\frac{dx}{dt} = -v$.
Therefore,$\varepsilon = -B \cdot l \cdot (-v) = B \cdot l \cdot v$.
Thus,the induced $emf$ is $\varepsilon = Blv$.