ic flux through the loop.

The magnetic flux $$\Phi$$ through the loop is given by:

$$\Phi = B \cdot A$$

where

$$B$$ is the magnetic field

and

$$A$$ is the area of the loop.

The area $$A$$ of a circular loop with radius $$r$$ is given by:

$$A = \pi r^2$$

The rate of change of the area $$A$$ with respect to time $$t$$ is:

$$\frac{dA}{dt} = \pi \frac{d}{dt} (r^2) = \pi \cdot 2r \cdot \frac{dr}{dt}$$

We are given:

$$\frac{dr}{dt} = -2 \mathrm{~mm} \mathrm{~s}^{-1} = -2 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-1}$$

Substituting these values in, we get:

$$\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt}$$

$$\frac{dA}{dt} = \pi \cdot 2 \cdot 0.04 \cdot -2 \times 10^{-3}$$

$$\frac{dA}{dt} = \pi \cdot 2 \cdot 0.04 \cdot -2 \cdot 10^{-3}$$

$$\frac{dA}{dt} = -1.6 \times 10^{-4} \pi \mathrm{~m}^2 \mathrm{~s}^{-1}$$

The induced emf $$\varepsilon$$ is given by Faraday's law of induction:

$$\varepsilon = - \frac{d\Phi}{dt}$$

Since$$ \Phi = B \cdot A$$, we have:

$$\varepsilon = - B \cdot \frac{dA}{dt}$$

Substituting the values of $$B$$ and $$\frac{dA}{dt}$$, we get:

$$\varepsilon = -0.125 \mathrm{~T} \cdot -1.6 \times 10^{-4} \pi \mathrm{~m}^2 \mathrm{~s}^{-1}$$

$$\varepsilon = 0.125 \cdot 1.6 \times 10^{-4} \pi$$

$$\varepsilon = 2 \times 10^{-5} \pi \mathrm{~V}$$

Since the question asks for the answer in $$\mu V$$ (microvolts), we convert the emf:

$$\varepsilon = 20 \pi \mu \mathrm{V}$$

Thus, the correct answer is:

Option B: $$20 \pi \mu V$$

Electromagnetic Induction

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