#rotating rod

Self-Induction and Self-Inductance

A rod rotates at 10 rad/s in a 0.5 T field. If the length from the axis to the tip is 0.6 m, what is the emf induced?

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. ε = (1/2) B ω R² . ε = (1/2) × 0.5 × 10 × (0.6)² = 0.9 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 I², result 0.9 V follows, reflecting Faraday's law and Lenz's opposition.

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

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AC Generator, Back EMF and Time Duration of EMF

A rod rotates at 25 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.8 m, what is the emf induced?

**Back emf** in motor opposes applied voltage, e_b = N B A ω sin ωt, reduces net current, at start ω=0 e_b=0 current large, as speed increases e_b increases limiting current, power conversion mechanical, principle of motor and generator reciprocity. ε = (1/2) B ω R² . ε = (1/2) × 0.3 × 25 × (0.8)² = 2.4 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 I², result 2.4 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Rotational EMF and AC Generator

A rod rotates at 20 rad/s in a 0.4 T field. If the length from the axis to the tip is 0.5 m, what is the emf induced?

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. ε = (1/2) B ω R² . ε = (1/2) × 0.4 × 20 × (0.5)² = 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) and U = ½ L I², result 5 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Rotational EMF and AC Generator

A rod rotates at 18 rad/s in a 0.4 T field. If the length from the axis to the tip is 0.7 m, what is the emf induced?

**AC generator** principle same as rotating coil, N=200 turns A=0.04 m² B=0.1 T f=50 Hz ω=2π×50=314 rad/s, e₀= N B A ω =200×0.1×0.04×314=251.2 V, emf e= e₀ sin ωt, frequency equals rotation frequency, maximum when plane parallel to field. ε = (1/2) B ω R² . ε = (1/2) × 0.4 × 18 × (0.7)² = 1.764 V ≈ 1.76 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 I², result 1.76 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Rotational EMF and AC Generator

A rod rotates at 16 rad/s in a 0.6 T field. If the length from the axis to the tip is 0.5 m, what is the emf induced?

**AC generator** principle same as rotating coil, N=200 turns A=0.04 m² B=0.1 T f=50 Hz ω=2π×50=314 rad/s, e₀= N B A ω =200×0.1×0.04×314=251.2 V, emf e= e₀ sin ωt, frequency equals rotation frequency, maximum when plane parallel to field. ε = (1/2) B ω R² . ε = (1/2) × 0.6 × 16 × (0.5)² = 1.2 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 I², result 1.2 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Rotational EMF and AC Generator

A rod rotates at 22 rad/s in a 0.4 T field. If the length from the axis to the tip is 0.6 m, what is the emf induced?

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. ε = (1/2) B ω R² . ε = (1/2) × 0.4 × 22 × (0.6)² = 1.584 V ≈ 1.58 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 I², result 1.58 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Magnetic Flux and Faraday's Laws of Induction

A conducting rod is rotated about one end in a uniform magnetic field. The emf induced between the ends is due to what f

**Magnetic flux** Φ = B·A = B A cosθ, B magnetic field (T), A area (m²), θ angle between B and normal to area, unit Wb = T·m², Faraday's law induced emf e = -N dΦ/dt, N turns, negative sign Lenz's law indicating opposition, magnitude |e| = N |ΔΦ/Δt|, for 100 turns ΔΦ=0.03 Wb Δt=0.06 s e=100×0.03/0.06=50 V. The rotation causes charges to move through the field, experiencing a magnetic force (Lorentz force) that separates them, inducing an emf along the rod. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω

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

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Rotational EMF and AC Generator

A rod rotates at 30 rad/s in a 0.6 T field. If the length from the axis to the tip is 0.8 m, what is the 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. ε = (1/2) B ω R² . ε = (1/2) × 0.6 × 30 × (0.8)² = 5.76 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L

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

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Motional EMF - Rod and Rectangular Loop

A rod rotates at 12 rad/s in a 0.5 T field. If the length from the axis to the tip is 0.9 m, what is the emf induced?

**Motional emf** for rod length l moving with velocity v perpendicular to uniform field B, e = B l v (V), B in T, l in m, v in m/s, direction given by right-hand rule, positive end where positive charges accumulate due to q v×B force. For l=0.9 m, v=1.2 m/s, B=0.5 T, e=0.5×0.9×1.2=0.54 V. ε = (1/2) B ω R² . ε = (1/2) × 0.5 × 12 × (0.9)² = 2.43 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 =

Ref: NCERT > Physics Book > Electromagnetic Induction > Motional EMF - Rod and Rectangular Loop

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Motional EMF - Rod and Rectangular Loop

A rod rotates at 25 rad/s in a 0.5 T field. If the length from the axis to the tip is 0.7 m, what is the emf induced?

**Loop sides 35 cm and 15 cm** moving out B=0.8 T v=1.5 m/s perpendicular to shorter side 15 cm, so cutting side =35 cm=0.35 m? Actually motion perpendicular to shorter side means longer side cuts, e= B×(long side)×v =0.8×0.35×1.5=0.42 V, illustrating motional emf e = B L v. ε = (1/2) B ω R² . ε = (1/2) × 0.5 × 25 × (0.7)² = 3.0625 V ≈ 3.06 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 =

Ref: NCERT > Physics Book > Electromagnetic Induction > Motional EMF - Rod and Rectangular Loop

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