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Q.

A uniform conducting ring of mass m = 2 kg, radius r = 2m and resistance 8Ω   is kept on smooth, horizontal surface (xz-plane). A time varying magnetic field  B=(i^+t2j^) Tesla is uniformly present in the region, where t is time in second and take vertical as y-axis. (Take  π2=10). Then   

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a

Heat generated through the ring till the instant when the ring start toppling is 40/3 Joule. 

b

Time when ring starts toppling is 1 sec.  

c

Heat generated through the ring till the instant when the ring start toppling is 80/3 Joule.

d

Time when ring starts toppling is 3/4 sec.

answer is A, D.

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Detailed Solution

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ϕ=BA=(t2)(πr2)ε=dϕdt=2πr2ti=εR=2πr2tRFor topplingτ=|M×B|mgr(iπr2)1mgr2πr2tRπr2mgrtmgR2π2r3=1  secHeat=i2Rdt=(2πr2tR)2dtRHeat=4π2r4R 01t2dt=803  Joule

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