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PhysicsEMI/ACSelf and mutual inductorHard2 minPYQ_2020
PhysicsHardnumerical

A non–conducting ring of radiusRhaving uniformly distributed chargeQstarts rotating aboutx–x'axis passing through diameter with an angular accelerationα, as shown in the figure. Another small conducting ring having radiusaa≪Ris kept fixed at the centre of bigger ring is such a way that axisxx'is passing through its centre and perpendicular to its plane. If the resistance of small ring isr=1 Ω,find the induced current in it inampere.
(Givenq=16×102μ0 C,R=1 m, a = 0.1 m,α=8 rad s-2)

Answer:
8.00
Solution:



dq=q2πR.Rdθ=q2π.dθ
di=dqT=qdθω2π2π
di=qω4π2.dθ

The magnetic field at O due to elementary ring is

dB=μ0di(Rsin⁡θ)22R3
∫dB=∫0πμ0sin2⁡θ2Rqω4π2dθ
B=μ0qω16πR

The flux of the magnetic field through the small ring

ϕ=Bπa2
ϕ=πa2.μ0qω16πR
ϕ=μ0qωa216R
|ε|=dϕdt
|ε|=μ0qa216Rα .
=8 volt
i=81=8 A.

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Self and mutual inductor
⏱ 2mℹ️ Source: PYQ_2020

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