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Electromagnetic Induction question

2010 · Q147
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Electromagnetic Induction question

2010 · Q147

NEETPhysicsElectromagnetic InductionMCQ+4 / −1
A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s−-−1. The induced emf when the radius is 2 cm, is
  1. A
    2πμV2\pi \mu V2πμV
  2. B
    πμV\pi \mu VπμV
  3. C
    π2μV{\pi \over 2}\mu V2π​μV
  4. D
    2μ\muμVVV
View written solutionFree

Correct answer: B

Constant rate at which radius of the loop shrinks

drdt=1×10−3{{dr} \over {dt}} = 1 \times {10^{-3}}dtdr​=1×10−3 m s-1

Magnetic flux linked with the loop is

ϕ\phi ϕ = BAcosϕ\phi ϕ = B(π\pi πr2)cos0o = Bπ\pi πr2

The magnitude of the induced emf is

|e| = dϕdt=ddt(Bπr2)=Bπ2rdrdt{{d\phi } \over {dt}} = {d \over {dt}}\left( {B\pi {r^2}} \right) = B\pi 2r{{dr} \over {dt}}dtdϕ​=dtd​(Bπr2)=Bπ2rdtdr​

= 0.025 ×\times× π\pi π ×\times× 2 ×\times× 2 ×\times× 10-2 ×\times× 1 ×\times× 10-3

= π\pi π ×\times× 10-6 V

= πμV\pi \mu VπμV

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