#moving loop

AC Generator, Back EMF and Time Duration of EMF

A loop of 0.2 m × 0.1 m moves out of a 0.5 T field at 2 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 = 0.1 m. t = (0.1/2) = 0.05 s . Using Φ = B A cosθ, e = -N dΦ/dt = -N A

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

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Motional EMF - Rod and Rectangular Loop

A loop of 0.4 m × 0.18 m moves out of a 0.5 T field at 1.8 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 = 0.18 m. t = (0.18/1.8) = 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

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

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AC Generator, Back EMF and Time Duration of EMF

A loop moves into a 0.3 T field at 1 m/s. If the length perpendicular to velocity is 0.1 m, how long does the emf last?

**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. Time = distance/velocity, where distance = width of loop along motion. Assume width = 0.1 m (typical NEET assumption). t = (0.1/1) = 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

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Motional EMF - Rod and Rectangular Loop

A loop of 0.25 m × 0.15 m moves out of a 0.2 T field at 1.5 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 = 0.15 m. t = (0.15/1.5) = 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

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

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