1. Use Faraday’s law

   The induced emf is

   $$\varepsilon = \left|\frac{d\Phi_B}{dt}\right|$$

   where magnetic flux through the loop is

   $$\Phi_B = BA$$

   as long as the part of the loop inside the magnetic field has area $A$.

2. Loop moving out with constant speed

   Let the rectangular loop have side length $\ell$ perpendicular to the direction of motion, and let it move with constant speed $v$ out of the uniform magnetic field region.

   While the loop is partially inside the field, the area inside the field decreases linearly with time:

   $$A(t) = \ell x(t)$$

   where $x(t)$ is the length of the loop still inside the field.

   Since the loop moves with constant speed,

   $$x(t) = x_0 - vt$$

   so

   $$A(t) = \ell(x_0-vt)$$

3. Flux variation

   Therefore,

   $$\Phi_B = B\ell(x_0-vt)$$

   Differentiating,

   $$\frac{d\Phi_B}{dt} = -B\ell v$$

   Hence the magnitude of induced emf is

   $$\varepsilon = B\ell v$$

   which is constant while the loop is partially inside the magnetic field.

4. Graph of $\varepsilon$ vs $t$

   So during the interval when the loop is leaving the field, the induced emf has a constant non-zero value.

   Before the loop starts leaving and after it completely leaves, flux does not change, so emf is zero. Thus the graph is a rectangular pulse: zero, then constant, then zero.

5. Matching with options

   The option representing a constant magnitude of induced emf during the partial خروج interval is Option B.

Conclusion: