1. Magnetic flux through the loop

Since the plane of the circular loop is perpendicular to the uniform magnetic field, the angle between the field and area vector is $0^\circ$. Hence,

$\Phi = BA = B\pi r^2$

2. Induced emf

By Faraday’s law, the magnitude of induced emf is

$\mathcal{E} = \left|\frac{d\Phi}{dt}\right| = \left|\frac{d}{dt}(B\pi r^2)\right|$

Since $B$ is constant,

$\mathcal{E} = B\left|\frac{d}{dt}(\pi r^2)\right| = B\cdot 2\pi r\left|\frac{dr}{dt}\right|$

3. Substitute the given values

Given:

• $B = 0.8\,\text{T}$
• $r = 10\,\text{cm} = 0.10\,\text{m}$
• shrinking rate $= 2\,\text{cm s}^{-1} = 0.02\,\text{m s}^{-1}$

So,

$\mathcal{E} = 0.8 \times 2\pi \times 0.10 \times 0.02$

$\mathcal{E} = 0.0032\pi\,\text{V}$

Using $\pi \approx 3.14$,

$\mathcal{E} \approx 0.0032 \times 3.14 = 0.010048\,\text{V}$

4. Convert to millivolts

$0.010048\,\text{V} = 10.048\,\text{mV}$

Thus, the induced emf is approximately

$\boxed{10\,\text{mV}}$

5. Comparison with stored answer

Stored correct answer = $10$

Our derived answer = $10$

They match.