In the figure as shown, a part of the wire belonging to the loop PQRS is bent into a semicircle of radius r = 0.2 m. Initially, the plane of the loop is perpendicular to the magnetic field B = 0.5 T. Now, the wire PQ is rotated from the starting position as shown in the figure to the final position attained after half a rotation. The change in the flux linked with the loop PQRS will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.02π Wb
The magnetic flux through the loop is \( \Phi = BA \), where A is the area enclosed by the loop and B is perpendicular to its plane. Only the semicircular part moves, so only its contribution to the area changes.
In the starting position the semicircle bulges outward, so its area is added to the area of the rectangle. After half a rotation about PQ it lies in the same plane but bulges inward, so its area is now subtracted.
\( \displaystyle A_{\mathrm{initial}} = A_{\mathrm{rect}} + \frac{\pi r^{2}}{2} \)
\( \displaystyle A_{\mathrm{final}} = A_{\mathrm{rect}} - \frac{\pi r^{2}}{2} \)
\( \displaystyle \Delta A = \pi r^{2} \)
\( \displaystyle \Delta\Phi = B\,\pi r^{2} \)
\( \displaystyle \Delta\Phi = 0.5\,\mathrm{T} \times \pi \times (0.2\,\mathrm{m})^{2} \)
\( \displaystyle \Delta\Phi = 0.02\pi\,\mathrm{Wb} \)