1. The problem statement, all variables and given/known data Charge Q is uniformly distributed on the rim of a thin insulating disc of mass m, radius R, which is initially at rest and placed on a smooth horizontal surface. What will be the angular velocity of the disk if a magnetic field B, perpendicular to the plane of the disk, is switched on? 2. Relevant equations ∫Eind.dl = -dφ/dt 3. The attempt at a solution 2πR. Eind = (B.πR2)/Δt E=BR/2Δt QE/m = QBR/2Δtm QE/m is acceleration. a = Rα (α is the angular acceleration) α = QB/2Δtm ω = ω0 + αΔt ω = QB/2Δtm × Δt = QB/2m However, the answer is QB/m. So, I tried another approach. 2πR. Eind = (B.πR2)/Δt E=BR/2Δt F=QE= QBR/2Δt τ=Iα QBR2/2Δt = (mR2)/2 × α From here, α = QB/Δtm and ω = QB/Δtm × Δt = QB/m And this is the correct answer. What is the problem with the first answer???