Question 1

The induced emf in the loop is given by Faraday's Law: $E = -\frac{d\Phi}{dt} = -B \frac{dA}{dt}$. Since the loop moves with a constant velocity $v$ along the $x$-axis, we can write $dx = v dt$, thus $E = -Bv \frac{dA}{dx}$, where $A(x)$ is the area of the loop that overlaps with the magnetic field region.

From the geometry provided, the $x$-axis points downwards. The magnetic field region is a triangle with its vertex at $x=L$ (relative to the starting position of $P$).

1. For $0 \le x \le L$: The loop enters the field vertex-first. The height of the triangular portion of the loop inside the field is $x$. The area of this portion is $A(x) = \frac{x^2}{\sqrt{3}}$. The induced emf is:

$E = -Bv \frac{d}{dx} \left( \frac{x^2}{\sqrt{3}} \right) = -\frac{2Bvx}{\sqrt{3}}$

This is a linear function of $x$ with a negative slope, going from $0$ at $x=0$ to $-\frac{2BvL}{\sqrt{3}}$ at $x=L$.

2. For $L \le x \le 2L$: The loop starts exiting the field region. The portion of the loop remaining in the field is a triangle of height $(2L-x)$. Its area is $A(x) = \frac{(2L-x)^2}{\sqrt{3}}$. The induced emf is:

$E = -Bv \frac{d}{dx} \left( \frac{(2L-x)^2}{\sqrt{3}} \right) = -Bv \left( \frac{2(2L-x)(-1)}{\sqrt{3}} \right) = \frac{2Bv(2L-x)}{\sqrt{3}}$

At $x=L$, $E = \frac{2BvL}{\sqrt{3}}$. At $x=2L$, $E = 0$. This segment also has a negative slope ($dE/dx = -2Bv/\sqrt{3}$).

The resulting graph shows a linear decrease to a negative peak at $x=L$, a discontinuous jump to a positive peak at $x=L$, and then a linear decrease to zero at $x=2L$. This behavior corresponds exactly to the graph shown in Option (A).