A conducting ring of mass $m=\pi kg$  and radius  $R=\frac{1}{2}m$ is kept on a flat horizontal surface (xy plane). A uniform magnetic field is switched on in the region which changes with time (t) as  $\overrightarrow{B}=(2\widehat{j}+t^2\widehat{k})T$. Resistance of the ring is  $r=\pi \Omega$  and  $g=10 m{s}^{-2}.$ Calculate the time t in seconds at which the ring begins to topple.

 

Flux through the ring is  $\varphi =\overrightarrow{B}.\overrightarrow{A}=(2\widehat{j}+t^2\widehat{k}).\left(\pi R^2\widehat{k}\right)=\pi R^2{t}^2$
 Induced emf   $\left|\epsilon \right|=\frac{d\varphi }{dt}=2\pi R^{2}t$
If induced electric field is  $E_{in}$ then    $E_{in}2\pi R=\epsilon$
$E_{in}2\pi R=2\pi R^{2}t$    
$\therefore E_{in}=Rt\mathrm{.................}\left(i\right)$
Current in the ring is  $i=\frac{\epsilon }{r}=\frac{2\pi R^{2}t}{r}$  [Clockwise]
Magnetic dipole moment  $\overrightarrow{\mu }=\left(\pi R^{2}i\right)(-\widehat{k})=\frac{2{\pi }^2{R}^{4}t}{r}(-\widehat{k})$
Torque on the ring   $\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}=\frac{4{\pi }^2{R}^{2}t}{r}\left(\widehat{i}\right)$
  
The ring will begin to topple when magnetic torque exceeds the torque due to weight about an axis tangential to the ring and parallel to  x axis. In the Figure the weight (W) of the ring is acting into the plane of the Figure. 
$\frac{4{\pi }^2{R}^{4}t}{r}=mgR \Rightarrow t_0=\frac{mg.r}{4{\pi }^2{R}^3}=\frac{\pi \times 10\times \pi }{4{\pi }^2\times {\left(\frac{1}{2}\right)}^3}=20\sec$