#moving rod

Self-Induction and Self-Inductance

A rod of length 0.6 m moves at 2.5 m/s in a 0.25 T field perpendicular to its length. What is the induced emf?

**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. ε = B l v = 0.25 × 0.6 × 2.5 = 0.375 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt

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

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Motional EMF - Rod and Rectangular Loop

A rod of length 0.3 m moves at 4 m/s in a 0.25 T field perpendicular to its length. What is the induced emf?

**Rectangular loop moving out of field** emf e = B l v, l side perpendicular to motion cutting field lines, e constant while partially in field, zero when fully out, duration t = (side parallel to motion)/v. For 0.38×0.55 m loop B=0.65 T v=0.7 m/s along shorter side 0.38 m, l=0.55 m (side cutting), e=0.65×0.55×0.7=0.25 V, lasts t=0.38/0.7=0.54 s. ε = B l v = 0.25 × 0.3 × 4 = 0.3 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 > Motional EMF - Rod and Rectangular Loop

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Motional EMF - Rod and Rectangular Loop

A rod of length 0.7 m moves at 2 m/s in a 0.3 T field perpendicular to its length. What is the induced emf?

**Rectangular loop moving out of field** emf e = B l v, l side perpendicular to motion cutting field lines, e constant while partially in field, zero when fully out, duration t = (side parallel to motion)/v. For 0.38×0.55 m loop B=0.65 T v=0.7 m/s along shorter side 0.38 m, l=0.55 m (side cutting), e=0.65×0.55×0.7=0.25 V, lasts t=0.38/0.7=0.54 s. ε = B l v = 0.3 × 0.7 × 2 = 0.42 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 > Motional EMF - Rod and Rectangular Loop

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Motional EMF - Rod and Rectangular Loop

A rod of length 0.4 m moves at 2 m/s in a 0.3 T field perpendicular to its length. What is the induced emf?

**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 = B l v = 0.3 × 0.4 × 2 = 0.24 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 0.24 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Motional EMF - Rod and Rectangular Loop

A rod of length 0.9 m moves at 1.2 m/s in a 0.5 T field perpendicular to its length. What is the induced emf?

**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 = B l v = 0.5 × 0.9 × 1.2 = 0.54 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

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

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