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 = B A = ( t 2 ) ( π r 2 ) ⇒ ε = d ϕ d t = 2 π r 2 t ⇒ i = ε R = 2 π r 2 t R F o r t o p p l i n g τ = | M → × B → | ≥ m g r ⇒ ( i ⋅ π r 2 ) ⋅ 1 ≥ m g r ⇒ 2 π r 2 t R ⋅ π r 2 ≥ m g r ⇒ t ≥ m g R 2 π 2 r 3 = 1 sec H e a t = ∫ i 2 R d t = ∫ ( 2 π r 2 t R ) 2 d t ⋅ R ⇒ H e a t = 4 π 2 r 4 R 0 1 t 2 d t = 80 3 J o u l e

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

Time when ring starts toppling is 1 sec.

Time when ring starts toppling is 3/4 sec.

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

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

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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