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#rotating disc

2 public questions tagged with this topic.

A conducting disc rotates in a uniform magnetic field parallel to its axis. The induced emf between the center and rim a

**Self-induction** emf induced in coil due to change in its own current, e = -L dI/dt, L self-inductance (H), L = μ₀ N² A / l for solenoid, N turns, A area (m²), l length (m), μ₀=4π×10⁻⁷ H/m. For solenoid 650 turns/m means n=650, A=0.014 m², L = μ₀ n² A l? Actually per unit length? For length l, N=n l, L= μ₀ n² A l, if l=1 m, L=4π×10⁻⁷×650²×0.014=7.43×10⁻³ H, dI/dt=(3-6)/0.25=-12 A/s, e= -L×(-12)=0.089 V. Rotation causes radial charge separation via the magnetic force ( F = q v × B ), inducing an emf from the center to the rim. Using Φ =

Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance

A metal disc rotates in a uniform magnetic field perpendicular to its plane. The induced emf between the center and the

**Flux change example** coil 150 turns area 0.06 m² B 0.14 T drops to zero in 0.3 s, ΔΦ = B A =0.14×0.06=0.0084 Wb per turn, ΔΦ/Δt=0.028 Wb/s, e=150×0.028=4.2 V, illustrating calculation from B and area. The rotation causes charges in the disc to move through the magnetic field, experiencing a Lorentz force that separates them radially, inducing an emf from center to rim. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L I², result Lorentz force on

Ref: NCERT > Physics Book > Electromagnetic Induction > Magnetic Flux and Faraday's Laws of Induction