#Faraday's law

Motional EMF - Rod and Rectangular Loop

A rectangular loop of sides 25 cm and 10 cm moves out of a 0.5 T field at 1.5 m/s perpendicular to the longer side. What

**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 , l = 0.1 m . ε = 0.5 × 0.1 × 1.5 = 0.075 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

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Motional EMF - Rod and Rectangular Loop

A rectangular loop of 0.18 m × 0.3 m moves out of a 0.25 T field at 1 m/s along its shorter side. What is the 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 , l = 0.3 m . ε = 0.25 × 0.3 × 1 = 0.075 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,

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Induced EMF Due to Change in Magnetic Field

A coil of 140 turns and area 0.05 m² is in a 0.1 T field that drops to zero in 0.25 s. What is the induced emf?

**Uniform field change** in coil produces emf proportional to area and turns, for circular coil radius 0.16 m area πr²=0.0804 m², B 0.12 T deformed to wire in 0.6 s, ΔΦ=0.12×0.0804=0.00965 Wb, e=0.00965/0.6=0.0161 V, illustrating area change also induces emf. Δ Φ = B A = 0.1 × 0.05 = 0.005 Wb . ε = N (Δ Φ/Δ t) = 140 × (0.005/0.25) = 140 × 0.02 = 2.8 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 =

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Induced EMF Due to Change in Magnetic Field

A coil of 120 turns and area 0.025 m² is in a field that decreases from 0.08 T to 0 in 0.4 s. 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. Δ Φ = B A = 0.08 × 0.025 = 0.002 Wb . ε = N (Δ Φ/Δ t) = 120 × (0.002/0.4) = 120 × 0.005 = 0.6 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 =

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Magnetic Flux and Faraday's Laws of Induction

A coil is placed in a time-varying magnetic field. The direction of the induced current is determined by which principle

**Faraday's first law** emf induced when flux linking coil changes, second law magnitude proportional to rate of change, e = -dΦ/dt, for N turns e = -N dΦ/dt, flux Φ = B A cosθ, change can be due to B change, A change, or θ change, all produce emf. Lenz’s law dictates that the induced current opposes the change in magnetic flux, determining its direction based on the field’s variation. 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

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Magnetic Flux and Faraday's Laws of Induction

A coil of 100 turns and area 0.03 m² is in a 0.08 T field that drops to zero in 0.2 s. What is the induced emf?

**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. Δ Φ = B A = 0.08 × 0.03 = 0.0024 Wb . ε = N (Δ Φ/Δ t) = 100 × (0.0024/0.2) = 100 × 0.012 = 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

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

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Magnetic Flux and Faraday's Laws of Induction

A conducting loop is expanded in a uniform magnetic field. The induced emf is caused by what physical process?

**Faraday's first law** emf induced when flux linking coil changes, second law magnitude proportional to rate of change, e = -dΦ/dt, for N turns e = -N dΦ/dt, flux Φ = B A cosθ, change can be due to B change, A change, or θ change, all produce emf. Expanding the loop increases the magnetic flux through it, and the rate of this flux change induces an emf as per Faraday’s law. 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

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Motional EMF - Rod and Rectangular Loop

A rectangular loop of sides 10 cm and 5 cm moves out of a 0.2 T magnetic field at 2 m/s perpendicular to the longer side

**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. For motional emf: ε = B l v , where l is the length perpendicular to velocity. Here, B = 0.2 T , l = 0.05 m , v = 2 m/s . ε = 0.2 × 0.05 × 2 = 0.02 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A

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Magnetic Flux and Faraday's Laws of Induction

A circular loop is placed in a uniform magnetic field, and the field strength is increased steadily. What determines the

**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. According to Faraday’s law, the induced emf depends on the rate of change of magnetic flux, which increases with the rate at which the magnetic field strength changes. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v =

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

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Rotational EMF and AC Generator

A coil of 200 turns rotates at 90 rad/s in a 0.04 T field. If the area is 0.015 m², what is the maximum emf?

**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. ε₀ = N B A ω = 200 × 0.04 × 0.015 × 90 = 10.8 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀

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

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Magnetic Flux and Faraday's Laws of Induction

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

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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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