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#motional emf

23 public questions tagged with this topic.

Self-Induction and Self-Inductance

A conducting rod moves perpendicular to a uniform magnetic field with constant velocity. What is true about the induced

**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. The induced emf ( ε = B l v ) remains constant because the velocity, magnetic field, and rod length are constant, leading to a steady flux change rate.

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

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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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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

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AC Generator, Back EMF and Time Duration of EMF

A loop of 0.45 m × 0.2 m moves out of a 0.6 T field at 1.5 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 = 0.2 m. t = (0.2/1.5) = 0.1333 s ≈ 0.13 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

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

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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of sides 30 cm and 12 cm moves out of a 0.6 T field at 1 m/s perpendicular to the shorter side. What

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. ε = B l v , l = 0.3 m . ε = 0.6 × 0.3 × 1 = 0.18 V . Using Φ = B A cosθ, e = -N

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

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Induced EMF Due to Change in Magnetic Field

A conducting rod is moved perpendicular to a uniform magnetic field. The emf induced across its ends depends on which of

**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. The motional emf is given by ε = B l v , where B is the magnetic field strength, l is the length of the rod, and v is its velocity, all of which are critical factors. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt,

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

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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of sides 15 cm and 6 cm moves out of a 0.4 T field at 1.5 m/s perpendicular to the shorter side. What

**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 = B l v , l = 0.15 m . ε = 0.4 × 0.15 × 1.5 = 0.09 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 0.09 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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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

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Motional EMF - Rod and Rectangular Loop

A rectangular loop of sides 28 cm and 14 cm moves out of a 0.7 T field at 1.2 m/s perpendicular to the longer side. What

**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 = 0.14 m . ε = 0.7 × 0.14 × 1.2 = 0.1176 V ≈ 0.118 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 > Motional EMF - Rod and Rectangular Loop

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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

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

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Motional EMF - Rod and Rectangular Loop

A rectangular loop of sides 12 cm and 4 cm moves out of a 0.35 T field at 2 m/s perpendicular to the longer side. What i

**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 = 0.04 m . ε = 0.35 × 0.04 × 2 = 0.028 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt,

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

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Motional EMF - Rod and Rectangular Loop

A rectangular loop of sides 45 cm and 25 cm moves out of a 1.0 T field at 0.8 m/s perpendicular to the longer side. What

**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 = 0.25 m . ε = 1.0 × 0.25 × 0.8 = 0.2 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,

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

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