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

ϕ=BA=(t2)(πr2)⇒ε=dϕdt=2πr2t⇒i=εR=2πr2tRFor topplingτ=|M→×B→|≥mgr⇒(i⋅π r2)⋅1≥mgr⇒2πr2tR⋅πr2≥mgr⇒t≥mgR2π2r3=1 secHeat=∫i2Rdt=∫(2πr2tR)2dt⋅R⇒Heat=4π2r4R 01t2dt=803 Joule

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

ϕ=BA=(t2)(πr2)⇒ε=dϕdt=2πr2t⇒i=εR=2πr2tRFor topplingτ=|M→×B→|≥mgr⇒(i⋅π r2)⋅1≥mgr⇒2πr2tR⋅πr2≥mgr⇒t≥mgR2π2r3=1 secHeat=∫i2Rdt=∫(2πr2tR)2dt⋅R⇒Heat=4π2r4R 01t2dt=803 Joule

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

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

Time when ring starts toppling is 1 sec.

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

Time when ring starts toppling is 3/4 sec.

answer is A, D.

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Detailed Solution

$\begin{array}{l} ϕ = B A = (t^{2}) (π r^{2}) \\ \Rightarrow ε = \frac{d ϕ}{d t} = 2 π r^{2} t \\ \Rightarrow i = \frac{ε}{R} = \frac{2 π r^{2} t}{R} \\ F o r t o p p l i n g \\ τ = | \vec{M} \times \vec{B} | \geq m g r \\ \Rightarrow (i \cdot π r^{2}) \cdot 1 \geq m g r \\ \Rightarrow \frac{2 π r^{2} t}{R} \cdot π r^{2} \geq m g r \\ \Rightarrow t \geq \frac{m g R}{2 π^{2} r^{3}} = 1 \sec \\ H e a t = \int^{} i^{2} R d t = \int^{} {(\frac{2 π r^{2} t}{R})}^{2} d t \cdot R \\ \Rightarrow H e a t = \frac{4 π^{2} r^{4}}{R}_{0}^{1} t^{2} d t = \frac{80}{3} J o u l e \end{array}$