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Electromagnetic Induction question

2014 · Q127
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Electromagnetic Induction question

2014 · Q127

NEETPhysicsElectromagnetic InductionMCQ+4 / −1
A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure.

AIPMT 2014 Physics - Electromagnetic Induction Question 24 English
The potential difference developed across the ring when its speed is vvv, is
  1. A
    zero
  2. B
    Bvπr22{{Bv\pi {r^2}} \over 2}2Bvπr2​ and P is at higher potential
  3. C
    πrBV\pi rBVπrBV and R is at higher potential
  4. D
    2rBV and R is at higher potential
View written solutionFree

Correct answer: D

Potential difference that is developed across ring when its speed is v :


ε\varepsilon ε = B v (li – lf)

where, li – lf = displacement between end of semicircular ring = 2r

Hence, ε\varepsilon ε = Bv(2r) where R is at high potential.

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