Q140 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
A wire '$l$' m long bent into a circular loop is placed perpendicular to the magnetic field of flux density $B \text{ Wbm}^{-2}$. Within 0.1sec, the loop is changed into a square of side '$a$' m and flux density is doubled. The value of e.m.f. induced is
- A$\dfrac{5 B\left(8 \pi a^2-l^2\right)}{2 \pi}$
- Correct answerB$\dfrac{-5 B\left(8 \pi a^2-l^2\right)}{2 \pi}$
- C$\dfrac{-5 B\left(8 \pi \mathrm{a}^2-\mathrm{l}^2\right)}{2}$
- D$\dfrac{-5 B\left(8 a^2-l^2\right)}{2 \pi}$
Solution
Initial flux: $\Phi_1 = B \cdot \pi r^2 = B \cdot \pi \left(\dfrac{l}{2\pi}\right)^2 = \dfrac{Bl^2}{4\pi}$
Final flux: $\Phi_2 = 2B \cdot a^2 = 2Ba^2$
Induced e.m.f.: $e = -\dfrac{\Delta\Phi}{\Delta t} = -\dfrac{2Ba^2 - \dfrac{Bl^2}{4\pi}}{0.1}$
$e = -10B\left(\dfrac{8\pi a^2 - l^2}{4\pi}\right) = \dfrac{-5B(8\pi a^2 - l^2)}{2\pi}$
Final flux: $\Phi_2 = 2B \cdot a^2 = 2Ba^2$
Induced e.m.f.: $e = -\dfrac{\Delta\Phi}{\Delta t} = -\dfrac{2Ba^2 - \dfrac{Bl^2}{4\pi}}{0.1}$
$e = -10B\left(\dfrac{8\pi a^2 - l^2}{4\pi}\right) = \dfrac{-5B(8\pi a^2 - l^2)}{2\pi}$
Hence, the correct answer is option: 'B'.