COMEDK · Physics · 22. Electromagnetic Induction
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 \(\dfrac{5 B\left(8 \pi a^2-l^2\right)}{2 \pi}\)
- B \(\dfrac{-5 B\left(8 \pi a^2-l^2\right)}{2 \pi}\)
- C \(\dfrac{-5 B\left(8 \pi \mathrm{a}^2-\mathrm{l}^2\right)}{2}\)
- D \(\dfrac{-5 B\left(8 a^2-l^2\right)}{2 \pi}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{-5 B\left(8 \pi a^2-l^2\right)}{2 \pi}\)
Step-by-step Solution
Detailed explanation
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.: \(e = -\dfrac{\Delta\Phi}{\Delta t} = -\dfrac{2Ba^2 - \dfrac{Bl^2}{4\pi}}{0.1}\)…
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