Electro Magnetic Induction And Alternating Currents Question 514

Question: A uniform circular loop of radius a and resistance R is placed perpendicular to a uniform magnetic field B. One half of the loop is rotated about the diameter with angular velocity ω as shown in Fig. Then, the current in the loop is

Options:

A) πa2Bω4R , when θ is zero

B) πa2Bω2R , when θ is zero

C) zero, when θ=π/2

D) πa2Bω2R , when θ=π/2

Show Answer

Answer:

Correct Answer: D

Solution:

  • θ=ωt .

    Only half circular part will be involved in inducing emf, so effective area

    A=πa22

    ϕ=BA,cos,θ

    e=dϕdt

    =+BA,sin,θ(dθdt)

    ,e=Bπa22ω,sin,θ

    I=eR=Bπa2ω2R,sin,θ



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक