1. Given data

• Magnetic field in region $x>0$: $B = 1\,\text{T}$
• Radius of ring: $r = 1\,\text{m}$
• Speed of ring along $+x$ direction: $v = 1\,\text{m/s}$
• At $t=0$, centre is at $x=-1\,\text{m}$

We need induced emf at $t=1\,\text{s}$.

2. Position of the centre at $t=1\,\text{s}$

Since the ring moves with constant velocity $1\,\text{m/s}$,

$x_c = -1 + (1)(1) = 0$

So at $t=1\,\text{s}$, the centre of the ring is exactly on the boundary $x=0$.

3. Magnetic flux through the ring

Magnetic field exists only in the region $x>0$. Therefore, at this instant, exactly the right half of the ring lies inside the magnetic field.

Flux through the ring is

$\Phi = B \times (\text{area of ring in } x>0)$

At $t=1\,\text{s}$, area inside field = area of semicircle:

$A = \frac{\pi r^2}{2} = \frac{\pi}{2}$

But induced emf is not just from flux value; it is from rate of change of flux:

$\mathcal{E} = \left|\frac{d\Phi}{dt}\right| = B\left|\frac{dA}{dt}\right|$

4. Rate of change of area at the boundary position

As the circle enters the field region, the area inside the field changes at a rate equal to

$\frac{dA}{dt} = v \times (\text{length of chord cut by the boundary})$

Here the boundary is the line $x=0$. When the centre is at $x=0$, this line passes through the diameter of the circle, so chord length is the full vertical diameter:

$\text{chord length} = 2r = 2\,\text{m}$

Thus,

$\frac{dA}{dt} = v(2r) = 1 \times 2 = 2\,\text{m}^2/\text{s}$

Hence,

$\mathcal{E} = B\frac{dA}{dt} = 1 \times 2 = 2\,\text{V}$

5. Check options

• A: $1\,\text{V}$ ❌
• B: $2\pi\,\text{V}$ ❌
• C: $2\,\text{V}$ ✅
• D: $0\,\text{V}$ ❌

6. Final answer

The induced emf at $t=1\,\text{s}$ is

$\boxed{2\,\text{V}}$