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PhysicsEMI/ACSelf and mutual inductorHard2 minPYQ_2020
PhysicsHardnumerical

A non–conducting ring of radiusRhaving uniformly distributed chargeQstarts rotating aboutxx'axis passing through diameter with an angular accelerationα, as shown in the figure. Another small conducting ring having radiusaaRis 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(Rsinθ)22R3
dB=0πμ0sin2θ2Rqω4π2dθ
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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