#rotating conductor

AC Generator, Back EMF and Time Duration of EMF

A rod rotates at 15 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.4 m, what is the emf induced?

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. ε = (1/2) B ω R² . ε = (1/2) × 0.3 × 15 × (0.4)² = 0.36 V . Using Φ = B A cosθ, e = -N dΦ/dt =

Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF

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Lenz's Law, Eddy Currents and Applications

A wheel with 8 spokes of 0.55 m each rotates at 50 rpm in a 0.4 T field. What is the induced emf?

**Eddy currents** are circulating currents induced in bulk conductor by changing flux, oppose motion, cause damping, heating, energy loss, minimized by laminating core into thin sheets insulated, increasing resistance, reducing eddy current magnitude, used in induction heating and braking. ω = 2π × (50/60) = (5π/3) rad/s . ε = (1/2) B ω R² = (1/2) × 0.4 × (5π/3) × (0.55)² = 0.3166 V ≈ 0.317 V . 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

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Rotational EMF and AC Generator

A square loop of side 18 cm rotates at 10 rad/s in a 0.25 T field. What is the maximum emf induced?

**Rotational emf** when coil area A rotates with angular speed ω in uniform field B, flux Φ = B A cos ωt, emf e = -N dΦ/dt = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = ω/2π. For square side 22 cm area 0.0484 m² N=1 ω=14 rad/s B=0.15 T, e₀=1×0.15×0.0484×14=0.1016 V, sinusoidal. A = (0.18)² = 0.0324 m² . ε₀ = N B A ω = 1 × 0.25 × 0.0324 × 10 = 0.081 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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Induced EMF Due to Change in Magnetic Field

A wheel with 4 spokes of 0.5 m each rotates at 48 rpm in a 0.8 T field. What is the induced emf?

**Field decreasing to zero** induces emf trying to maintain field, current direction such that its field adds to original. For 150 turns area 0.06 m² B 0.14 T to zero in 0.3 s, e=150×0.06×0.14/0.3=4.2 V, as earlier, showing linear dependence on N, A, ΔB/Δt. ω = 2π × (48/60) = 1.6π rad/s . ε = (1/2) B ω R² = (1/2) × 0.8 × 1.6π × (0.5)² = 0.5024 V ≈ 0.5 V . 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)

Ref: NCERT > Physics Book > Electromagnetic Induction > Induced EMF Due to Change in Magnetic Field

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PHYSICS

A wheel with 6 spokes of 0.3 m each rotates at 40 rpm in a 0.6 T field. What is the induced emf?

Given: A wheel with 6 spokes of 0.3 m each rotates at 40 rpm in a 0.6 T field. What is the induced emf? These values define the system as per NCERT data. Formula: omega = 2π × 40/60 = 4π/3 rad/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 1/2 B omega R² = 1/2 × 0.6 × 4π/3 × (0.3)² = 0.07536 V approx 0.075 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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