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Question

A small equilateral triangular loop of side l=1 cm is kept inside a large circular loop of radius r such that both are coplanar and concentric. If current in the larger loop is as shown in the graph, then find the current induced in the smaller loop during the time instants given. Take anticlock wise current as positive.


List - 1 gives time instants and list - 2 gives induced current in the smaller loop. Take μ0π2rR=1 where R is resistance of small loop.
List -1List -2(I)0 sec(p)25μA(II)13sec(Q)12.5μA(III)32sec(R)0μA(IV)53sec(S)6.25μA(T)12.5μA(U)25μA

A
IU,IIT,IIIR,IVT
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B
IP,IIQ,IIIR,IVT
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C
IU,IIS,IIIR,IVT
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D
IP,IIQ,IIIT,IVR
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Solution

The correct option is A IU,IIT,IIIR,IVT
Φ=BA=μ0i2r34l2=μ0234l2r13sinωt
=dϕdt=μ08l2rωcosωtItriangle=R and ω=2πT=2π2
Itriangle=μ0l2π8rRcos ωt=μ0π2rR.l2cos ωt4=25cos ωt μA
Itriangle=25cosωt μA
at t=0,Itriangle=25μA clock wise
So at t=0,Itriangle=25μA
and at t=13,Itriangle=25cosπ3=12.5μA clockwise
So t=13Itriangle=12.5μA
at t=32Itriangle=0
At t=53,Itriangle=25cos5A3=12.5μA clockwise
so at t=53,Itriangle=12.5μA
IU,IIT,IIIR,IVT

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