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Electromagnetic Induction question

2019 · Shift 1 · Q42
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  5. /2019 · Shift 1 · Q42

Electromagnetic Induction question

2019 · Shift 1 · Q42

JEE AdvancedPhysicsElectromagnetic InductionMultiple correct+4 / −1
A conducting wire of parabolic shape, initially y = x2, is moving with velocity v=v0i^v = {v_0}\widehat iv=v0​i in a non-uniform magnetic field B=B0(1+(yL)β)k^B = {B_0}\left( {1 + {{\left( {{y \over L}} \right)}^\beta }} \right)\widehat kB=B0​(1+(Ly​)β)k, as shown in figure. If V0, B0, L and β\betaβ are positive constants and Δϕ\Delta \phiΔϕ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are JEE Advanced 2019 Paper 1 Offline Physics - Electromagnetic Induction Question 16 English
  1. A
    ∣Δϕ∣=43B0V0L\left| {\Delta \phi } \right| = {4 \over 3}{B_0}{V_0}L∣Δϕ∣=34​B0​V0​L for β\betaβ = 2
  2. B
    ∣Δϕ∣\left| {\Delta \phi } \right|∣Δϕ∣ remains the same if the parabolic wire is replaced by a straight wire, y =x initially, of length 2L\sqrt 2 L2​L
  3. C
    ∣Δϕ∣\left| {\Delta \phi } \right|∣Δϕ∣=12B0V0L{1 \over 2}{B_0}{V_0}L21​B0​V0​L for β\betaβ = 0
  4. D
    ∣Δϕ∣\left| {\Delta \phi } \right|∣Δϕ∣ is proportional to the length of the wire projected on the y-axis.
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