#induced voltage

Energy Stored in Inductor and Magnetic Energy

A solenoid’s current is switched off suddenly. What causes sensitive instruments nearby to be at risk of damage?

**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. The sudden drop in current induces a high emf due to self-inductance, which can generate significant currents or voltages in nearby circuits, risking damage. 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 > Energy Stored in Inductor and Magnetic Energy

View Details

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

View Details

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

View Details

AC Generator, Back EMF and Time Duration of EMF

A loop of 0.5 m × 0.22 m moves out of a 0.7 T field at 1.4 m/s along its longer side. How long does the emf last?

**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. Time = distance/velocity, distance = width along motion = 0.22 m. t = (0.22/1.4) = 0.1571 s ≈ 0.16 s . Using Φ = B A cosθ, e = -N dΦ/dt

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

View Details

Motional EMF - Rod and Rectangular Loop

A rectangular loop of sides 20 cm and 8 cm moves out of a 0.45 T field at 1.2 m/s perpendicular to the shorter side. Wha

**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 = 0.2 m . ε = 0.45 × 0.2 × 1.2 = 0.108 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

View Details

Motional EMF - Rod and Rectangular Loop

A rectangular loop of 0.22 m × 0.4 m moves out of a 0.35 T field at 0.5 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 = 0.4 m . ε = 0.35 × 0.4 × 0.5 = 0.07 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

View Details

Motional EMF - Rod and Rectangular Loop

A rectangular loop of 0.38 m × 0.55 m moves out of a 0.65 T field at 0.7 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 = 0.55 m . ε = 0.65 × 0.55 × 0.7 = 0.25025 V ≈ 0.25 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

View Details