#magnetic induction

Energy Stored in Inductor and Magnetic Energy

A wheel with 5 spokes of 0.4 m each rotates at 30 rpm in a 0.7 T field. What is the induced emf?

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. ω = 2π × (30/60) = π rad/s . ε = (1/2) B ω R² = (1/2) × 0.7 × π × (0.4)² = 0.1759 V ≈ 0.176 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

View Details

Mutual Induction and Mutual Inductance

A coil is wound tightly around a core material. If the current through it changes rapidly, the induced emf opposing this

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. This is self-induction, where a changing current in a coil induces an emf that opposes the change, proportional to the coil’s self-inductance. 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 Self-induction follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

View Details

Induced EMF Due to Change in Magnetic Field

A coil of 100 turns and area 0.02 m² is rotated at 50 Hz in a 0.05 T field. What is the maximum emf?

**Induced emf due to B change** e = -N A dB/dt, N turns, A area (m²), dB/dt rate of change of field (T/s). For 110 turns area 0.035 m² B 0.09 T to 0 in 0.5 s, dB/dt=0.18 T/s, e=110×0.035×0.18=0.693 V, direction opposes decrease via Lenz's law. ω = 2π v = 2π × 50 = 100π rad/s . ε₀ = N B A ω = 100 × 0.05 × 0.02 × 100π = 31.4 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,

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

View Details

AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of sides 40 cm and 20 cm moves out of a 0.9 T field at 1 m/s perpendicular to the longer side. What i

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. ε = B l v , l = 0.2 m . ε = 0.9 × 0.2 × 1 = 0.18 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 > AC Generator, Back EMF and Time Duration of EMF

View Details

Rotational EMF and AC Generator

A circular coil of radius 5 cm and 100 turns is rotated in a 0.1 T magnetic field at 50 rad/s. What is the maximum emf i

**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. Maximum emf: ε₀ = N B A ω . A = π r² = 3.14 × (0.05)² = 0.00785 m² . ε₀ = 100 × 0.1 × 0.00785 × 50 = 3.925 V ≈ 3.93 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,

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

View Details

Magnetic Flux and Faraday's Laws of Induction

A magnet is moved away from a coil, and the induced current flows clockwise as viewed from the magnet. What is the polar

**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. By Lenz’s law, the clockwise current opposes the decreasing flux by producing a south pole facing the receding north pole of the magnet, creating an attractive force. 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 > Magnetic Flux and Faraday's Laws of Induction

View Details

Rotational EMF and AC Generator

A circular coil of radius 4 cm and 250 turns rotates at 40 rad/s in a 0.06 T field. What is the maximum 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. A = π r² = 3.14 × (0.04)² = 0.005024 m² . ε₀ = N B A ω = 250 × 0.06 × 0.005024 × 40 = 3.0144 V ≈ 3.01 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

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

View Details

Motional EMF - Rod and Rectangular Loop

A loop of 0.32 m × 0.16 m moves out of a 0.35 T field at 2 m/s along its longer side. How long does the emf last?

**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. Time = distance/velocity, distance = width along motion = 0.16 m. t = (0.16/2) = 0.08 s . 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,

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

View Details

Motional EMF - Rod and Rectangular Loop

A loop of 0.35 m × 0.1 m moves out of a 0.25 T field at 2.5 m/s along its longer side. How long does the emf last?

**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. Time = distance/velocity, distance = width along motion = 0.1 m. t = (0.1/2.5) = 0.04 s . 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

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

View Details

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

View Details

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

View Details

Rotational EMF and AC Generator

A circular coil of radius 6 cm and 200 turns rotates at 30 rad/s in a 0.08 T field. What is the maximum 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. A = π r² = 3.14 × (0.06)² = 0.0113 m² . ε₀ = N B A ω = 200 × 0.08 × 0.0113 × 30 = 5.424 V ≈ 5.42 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) and

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

View Details