According to Faraday’s law of electromagnetic induction the induced emf is ε = – $\frac{d\phi}{dt}$ Thus a relation between electric field and rate of change of flux can be established. ε = - ∫ $\vec{E}\cdot\vec{dl}$ = - $\frac{d\phi}{dt}$ $\vec{E}$ exist along circumference of radius ‘a’ due to change in magnetic flux. E ∫ dl = - $\frac{d}{dt}(\pi a^2 B)$ , E × 2πa = - $\pi a^2 \frac{dB}{dt}$ E = - $\frac{a}{2} \frac{dB}{dt}$ ... (i) Linear charge density on rim is λ. So, total charge on rim Q = λ2πa ...(ii) Electric Force on the charge F = QE = $-\pi a^2 \lambda \frac{dB}{dt}$ ⇒ m $\frac{dv}{dt}$ = - $\pi a^2$ λ $\frac{dB}{dt}$ In terms of angular velocity v = Rω m $\frac{d}{dt}$ (Rω) = - $\pi a^2 \lambda \frac{dB}{dt}$ mR dω = - $\pi a^2 \lambda dB$ dω = - $\frac{\pi a^2 \lambda}{m R}$ dB Integrating both sides, ω = - $\frac{\pi a^2 \lambda B}{m R}$ As direction of angular velocity is along axis. $\omega^{\rightarrow}$ = - $\frac{\lambda a^2 \pi}{m R} \widehat{Bk}$