1. Magnetic flux through the loop

Since the plane of the circular loop is perpendicular to the magnetic field, the area vector is parallel to the field. So, $\Phi = BA$ where $A = \pi r^2$ Hence, $\Phi = B\pi r^2$

2. Induced emf

The magnitude of induced emf is $|\mathcal{E}| = \left|\frac{d\Phi}{dt}\right|$

Since $B$ is constant, $|\mathcal{E}| = B\frac{d}{dt}(\pi r^2)$ $|\mathcal{E}| = B \cdot 2\pi r \frac{dr}{dt}$

3. Substitute the given values

Given:

• $B = 0.4\,\text{T}$
• $r = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m}$
• $\dfrac{dr}{dt} = 1\,\text{mm/s} = 1 \times 10^{-3}\,\text{m/s}$

So, $|\mathcal{E}| = 0.4 \times 2\pi \times (2 \times 10^{-2}) \times (1 \times 10^{-3})$

4. Calculation

$|\mathcal{E}| = 0.4 \times 2\pi \times 2 \times 10^{-5}$ $|\mathcal{E}| = 1.6\pi \times 10^{-5}$

Using $\pi \approx 3.14$, $|\mathcal{E}| \approx 1.6 \times 3.14 \times 10^{-5}$ $|\mathcal{E}| \approx 5.024 \times 10^{-5}\,\text{V}$

5. Convert to microvolts

Since $1\,\mu\text{V} = 10^{-6}\,\text{V}$ we get $|\mathcal{E}| = 5.024 \times 10^{-5}\,\text{V} = 50.24\,\mu\text{V}$

For an integer answer, $\boxed{50}$

6. Comparison with stored answer

Stored correct answer = $50$

Our derived answer = $50$

So they agree.