A wire bent into a circular loop of radius r is placed with its plane perpendicular to a uniform magnetic field that changes at a constant rate, and an emf \( \varepsilon \) is induced in it. The same wire is now bent into a square loop and placed in the same way in the same field. The emf induced in the square loop is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \frac{\pi\varepsilon}{4} \)
By Faraday’s law, the emf induced in a loop equals the rate of change of the magnetic flux through it. With the plane of the loop perpendicular to the field, the flux is \( \Phi_B=BA \). So for the same rate of change of \( B \), the emf is proportional to the area of the loop.
The same wire gives both loops the same perimeter. For a square of side \( a \):
\( 4a=2\pi r \)
\( \displaystyle \Rightarrow\; a=\frac{\pi r}{2} \)
The area of the square loop is:
\( \begin{aligned} A_{\mathrm{sq}} &= a^{2} \\ &= \frac{\pi^{2}r^{2}}{4} \end{aligned} \)
The ratio of the two emfs equals the ratio of the areas:
\( \begin{aligned} \frac{\varepsilon_{\mathrm{sq}}}{\varepsilon} &= \frac{A_{\mathrm{sq}}}{\pi r^{2}} \\ &= \frac{\pi^{2}r^{2}/4}{\pi r^{2}} \\ &= \frac{\pi}{4} \end{aligned} \)
Therefore, the emf induced in the square loop is \( \pi\varepsilon/4 \). This is a little less than \( \varepsilon \), because for the same length of wire a circle encloses a larger area than a square.