1. Given data

• Side of square loop:
  $a = 15\text{ cm} = 0.15\text{ m}$
• Speed of loop:
  $v = 2\text{ cm/s} = 0.02\text{ m/s}$
• Width of magnetic field region:
  $L = 50\text{ cm} = 0.50\text{ m}$
• Front edge enters the magnetic field at $t=0$.

We need induced emf at $t=10\text{ s}$.

2. Find how far the loop has moved in 10 s

$x = vt = 2\times 10 = 20\text{ cm}$

So after 10 s, the front edge has moved 20 cm into the field.

3. Check whether the loop is entering, fully inside, or leaving

The loop side is $15\text{ cm}$.

• At $t=0$: front edge just enters.
• Time for the entire loop to enter field: $t_1 = \frac{15}{2} = 7.5\text{ s}$

So after $7.5\text{ s}$, the whole loop is completely inside the magnetic field.

Now, to start leaving the field, the front edge must reach the far boundary of the field region, i.e. travel $50\text{ cm}$.

• Time when front edge reaches far boundary: $t_2 = \frac{50}{2} = 25\text{ s}$

Thus:

• For $0 < t < 7.5\text{ s}$: flux increases, emf induced.
• For $7.5 < t < 25\text{ s}$: loop completely inside uniform field, flux constant, emf zero.
• For $25 < t < 32.5\text{ s}$: loop leaves field, emf induced again.

Since $t=10\text{ s}$ lies in the interval $7.5 < 10 < 25,$ the loop is fully inside the uniform magnetic field.

4. Induced emf

Induced emf is $\mathcal{E} = -\frac{d\Phi}{dt}$

When the loop is fully inside a uniform magnetic field, the magnetic flux through it is constant. Hence, $\frac{d\Phi}{dt}=0$

Therefore, $\mathcal{E}=0$

5. Evaluate options

• A: zero ✅
• B: 4.5 mV ❌
• C: 0.3 mV ❌
• D: 3 mV ❌

So the correct option is: $\boxed{\text{A: zero}}$

6. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They match.