A rectangular metallic loop is moving out of a uniform magnetic field region to a field free region with a constant speed. When the loop is partially inside the magnate field, the plot of magnitude of induced emf (ε) with time (t) is given by

Let the uniform magnetic field $$\mathbf{B}$$ be perpendicular to the plane of a rectangular loop of breadth $$b$$ (side perpendicular to the direction of motion) and length $$\ell$$ (side parallel to the direction of motion).
The loop is pulled out of the field with a constant speed $$v$$.

Magnetic flux linked with the loop is
$$\Phi = B \times \text{(area of loop that still lies inside the field)}$$

1. For $$t \lt t_1$$ : the whole loop is inside the field, so the enclosed area is constant ( $$= b\ell$$ ).
 $$\displaystyle \frac{d\Phi}{dt}=0 \;\Rightarrow\; \varepsilon = 0$$

2. From $$t_1$$ to $$t_2$$ : the front edge has crossed the field boundary, the rear edge is still inside.
 At an instant when the front edge has moved a distance $$x = v(t-t_1)$$ out of the field, the length still inside is $$\ell - x$$.
 Enclosed area $$A = b(\ell - x)$$, therefore
 $$\Phi = B b(\ell - x)$$

Using Faraday’s law $$\varepsilon = \left|\dfrac{d\Phi}{dt}\right|$$:
 $$\varepsilon = \left|B b \dfrac{d(-x)}{dt}\right| = B b v$$

• $$B$$, $$b$$ and $$v$$ are constants ⇒ $$\varepsilon$$ remains constant in this entire interval.
• The interval lasts until the rear edge reaches the boundary: $$x = \ell \Rightarrow t_2 - t_1 = \dfrac{\ell}{v}$$.

3. For $$t \gt t_2$$ : the whole loop is outside, flux is zero again, so $$\varepsilon = 0$$.

Hence the magnitude-time graph is
• zero up to $$t_1$$,
• a horizontal line of height $$B b v$$ from $$t_1$$ to $$t_2$$,
• zero after $$t_2$$.

This is a rectangular (top-hat) pulse, matching the sketch shown as Option D (image 4).
Therefore, the correct choice is Option D.