We start by noting that the induced emf (by Faraday’s law) is given by

$$ \varepsilon = -\frac{\Delta\Phi}{\Delta t}, $$

where the flux change

$$ \Delta\Phi = \Phi_{\text{final}} - \Phi_{\text{initial}}. $$

In our problem the magnetic field is perpendicular to the loop (so we don’t have to worry about any angle factors) and its value doubles when the loop is changed in 0.1 sec. Let’s set up the steps:

1. Define l as the length of the wire (in cm) and a as the side of the square (in cm) into which the wire is reformed. (In many problems the wire length l is used; here the given value “1 cm” is simply a particular case, so we write our answer in terms of l and a.)

2. When the wire is bent into a circle, its circumference is l. That means the radius r of the circle is

$$ r = \frac{l}{2\pi}. $$

Therefore, the area of the circular loop is

$$ A_{\text{circle}} = \pi r^2 = \pi\left(\frac{l}{2\pi}\right)^2 = \frac{l^2}{4\pi}. $$

3. When the loop is changed into a square of side a, its area is simply

$$ A_{\text{square}} = a^2. $$

(Note: In a typical wire‐reshaping problem the new shape’s perimeter would equal l so that a would be l/4. Here, however, the side a is given explicitly, so we use it as an independent parameter.)

4. The initial magnetic flux is that through the circular loop in a field B:

$$ \Phi_{\text{initial}} = B\,A_{\text{circle}} = B\,\frac{l^2}{4\pi}. $$

After 0.1 sec the loop is made into a square and, at the same time, the magnetic flux density is doubled to 2B. Therefore the final flux is

$$ \Phi_{\text{final}} = (2B)\,A_{\text{square}} = 2B\,a^2. $$

5. Thus the change in flux is

$$ \Delta\Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 2B\,a^2 - B\,\frac{l^2}{4\pi}. $$

6. Plug this into Faraday’s law with the time interval $ \Delta t = 0.1\,\text{sec} $:

$$ \varepsilon = -\frac{\Delta\Phi}{\Delta t} = -\frac{2B\,a^2 - \dfrac{B\,l^2}{4\pi}}{0.1}. $$

Multiplying numerator and denominator by 10 (since $1/0.1 = 10$) gives

$$ \varepsilon = -10B\left(2a^2 - \frac{l^2}{4\pi}\right). $$

7. To combine the terms inside the parentheses, rewrite $2a^2$ with a common denominator:

$$ 2a^2 = \frac{8\pi\,a^2}{4\pi}. $$

Thus we have

$$ \varepsilon = -10B\,\left(\frac{8\pi\,a^2 - l^2}{4\pi}\right) = -\frac{10B}{4\pi}\,(8\pi\,a^2 - l^2). $$

Simplify the factor $\frac{10}{4} = \frac{5}{2}$, so

$$ \varepsilon = -\frac{5B}{2\pi}\,(8\pi\,a^2 - l^2). $$

This is exactly the expression given in option C:

$$ \textbf{Option C:}\quad \varepsilon = \frac{-5B\,(8\pi a^2 - l^2)}{2\pi}. $$

Thus, the induced emf is

$$ \varepsilon = -\frac{5B\,(8\pi\,a^2 - l^2)}{2\pi}, $$

and the correct answer is Option C.

Electromagnetic Induction

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