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Electromagnetic Induction question

2016 · Q127
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Electromagnetic Induction question

2016 · Q127

NEETPhysicsElectromagnetic InductionMCQ+4 / −1
A uniform magnetic field is restricted within a region of rafius r. The magnetic field changes with time at a rate dB→dt{{d\overrightarrow B } \over {dt}}dtdB​. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure. Then the e.m.f. generated is

NEET 2016 Phase 2 Physics - Electromagnetic Induction Question 27 English
  1. A
    zero in loop 1 and zero in loop 2
  2. B
    −dB→dtπr2- {{d\overrightarrow B } \over {dt}}\pi {r^2}−dtdB​πr2 in loop 1 and −dB→dtπr2- {{d\overrightarrow B } \over {dt}}\pi {r^2}−dtdB​πr2 in loop 2
  3. C
    −dB→dtπR2- {{d\overrightarrow B } \over {dt}}\pi {R^2}−dtdB​πR2 in loop 1 and zero in loop 2
  4. D
    −dB→dtπr2- {{d\overrightarrow B } \over {dt}}\pi {r^2}−dtdB​πr2 in loop 1 and zero in loop 2
View written solutionFree

Correct answer: D

Emf generated in loop 1,

ε1=−dϕdt=−ddt(BA)=−A×dBdt{\varepsilon _1} = - {{d\phi } \over {dt}} = - {d \over {dt}}\left( {BA} \right) = - A \times {{dB} \over {dt}}ε1​=−dtdϕ​=−dtd​(BA)=−A×dtdB​

⇒\Rightarrow⇒ ε1=−πr2×dBdt{\varepsilon _1} = - \pi {r^2} \times {{dB} \over {dt}}ε1​=−πr2×dtdB​

Emf generated in loop 2,

ε2=−dϕdt=−ddt(BA)=−ddt(0×A){\varepsilon _2} = - {{d\phi } \over {dt}} = - {d \over {dt}}\left( {BA} \right) = - {d \over {dt}}\left( {0 \times A} \right)ε2​=−dtdϕ​=−dtd​(BA)=−dtd​(0×A) = 0

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