The uniform magnetic field perpendicular to the plane of a conducting ring of radius a…
Physics · JEE Advanced · NTA Exams — Electromagnetic Induction and Alternating Currents
The uniform magnetic field perpendicular to the plane of a conducting ring of radius a changes at the rate of α, then
- all the points on the ring are at the same potential
- the e.m.f. induced in the ring is πa2α
- electric field intensity E at any point on the ring is zero
- \(\mathrm{E}=\frac{\mathrm{a} \alpha}{2}\)
Answer
(B) the e.m.f. induced in the ring is π a 2 α, (D) E = a 2
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