#moving conductor

AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of 0.1 m × 0.2 m moves out of a 0.4 T field at 0.5 m/s along its shorter side. What is the emf?

**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 = 0.4 × 0.2 × 0.5 = 0.04 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

Induced EMF Due to Change in Magnetic Field

A rod of length 0.8 m moves at 1.5 m/s in a 0.4 T field perpendicular to its length. What is the induced 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. ε = B l v = 0.4 × 0.8 × 1.5 = 0.48 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 0.48 V follows, reflecting Faraday's law and Lenz's opposition.

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

View Details

Motional EMF - Rod and Rectangular Loop

A rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side. What is the 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 , l = 0.38 m . ε = 0.3 × 0.38 × 0.6 = 0.0684 V ≈ 0.068 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B

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

View Details

Motional EMF - Rod and Rectangular Loop

A rod of length 0.35 m moves at 2.5 m/s in a 0.2 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 = 0.2 × 0.35 × 2.5 = 0.175 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

View Details

AC Generator, Back EMF and Time Duration of EMF

A loop of 0.28 m × 0.14 m moves out of a 0.3 T field at 1.4 m/s along its longer side. How long does the emf last?

**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. Time = distance/velocity, distance = width along motion = 0.14 m. t = (0.14/1.4) = 0.1 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 U = ½ L I², result 0.1

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

View Details

Motional EMF - Rod and Rectangular Loop

A rod of length 0.5 m moves at 3 m/s in a 0.2 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 = 0.2 × 0.5 × 3 = 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

View Details

PHYSICS

A rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side. What is the emf?

Given: A rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side. What is the emf? These values define the system as per NCERT data. Formula: varepsilon = B l v, l = 0.38 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 0.3 × 0.38 × 0.6 = 0.0684 V approx 0.068 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.

View Details