In the figure shown, the conductor PQ of length \(l\) is moved from \(x = 0\) to \(x = b\) and then up to \(x = 2b\) with a constant velocity \(\vec{v}\). A uniform magnetic field \(\vec{B}\) is perpendicular to the plane of the paper and extends from \(x = 0\) to \(x = b\) and it is zero from \(x > b\). The magnitude of emf Induced in the conductor is

When the conductor PQ moves from \(x = 0\) to \(x = b\), it moves through a uniform magnetic field \(\vec{B}\) which is perpendicular to both its length \(l\) and velocity \(\vec{v}\).

The motional emf induced across the ends of the conductor is given by the formula:
\(E = B l v\)

Since \(B\), \(l\), and \(v\) are constant, the induced emf is constant and equal to \(B l v\) in the region \(0 \leq x \leq b\).

When the conductor moves from \(x = b\) to \(x = 2b\), the magnetic field is zero (\(B = 0\)).

Therefore, the induced emf in this region is zero.

Thus, the magnitude of emf induced is \(B l v\) for \(0 \leq x \leq b\).

Answer: \(B l v; 0 \leq x \leq b\)