A non–conducting ring of radius R having uniformly distributed charge Q starts rotating about x – x ' axis passing through diameter with an angular acceleration α , as shown in the figure. Another small conducting ring having radius a a ≪ R is kept fixed at the centre of bigger ring is such a way that axis xx ' is passing through its centre and perpendicular to its plane. If the resistance of small ring is r = 1 Ω ,

Step-by-step solution

d q = q 2 π R . R d θ = q 2 π . d θ d i = d q T = q d θ ω 2 π 2 π d i = q ω 4 π 2 . d θ The magnetic field at O due to elementary ring is d B = μ 0 d i ( R sin ⁡ θ ) 2 2 R 3 ∫ d B = ∫ 0 π μ 0 sin 2 ⁡ θ 2 R q ω 4 π 2 d θ B = μ 0 q ω 16 π R The flux of the magnetic field through the small ring ϕ = B π a 2 ϕ = π a 2 . μ 0 q ω 16 π R ϕ = μ 0 q