Chapter 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
Chapter 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
23.5 Electric Generators
- Calculate the emf induced in a generator.
- Calculate the peak emf which can be induced in a particular generator system.
Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Induced Emf and Magnetic Flux. We will now explore generators in more detail. Consider the following example.
The generator coil shown in [link] is rotated through one-fourth of a revolution (from θ=0ºθ=0º
to θ=90ºθ=90º
) in 15.0 ms. The 200-turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced?
Strategy
We use Faraday’s law of induction to find the average emf induced over a time ΔtΔt size 12{Δt} {}:
We know that N=200N=200 size 12{N=”200″} {} and Δt=15.0msΔt=15.0ms size 12{Δt=”15″ “.” 0`”ms”} {}, and so we must determine the change in flux ΔΦΔΦ size 12{ΔΦ} {} to find emf.
Solution
Since the area of the loop and the magnetic field strength are constant, we see that
Now, Δ(cosθ)=−1.0Δ(cosθ)=−1.0 size 12{Δ ( “cos”θ ) = – 1 “.” 0} {}, since it was given that θθ goes from 0º0º to 90º90º . Thus ΔΦ=−ABΔΦ=−AB size 12{ΔΦ= – ital “AB”} {}, and
The area of the loop is A=πr2=(3.14…)(0.0500m)2=7.85×10−3m2A=πr2=(3.14…)(0.0500m)2=7.85×10−3m2 size 12{A=πr rSup { size 8{2} } = ( 3 “.” “14” “.” “.” “.” ) ( 0 “.” “0500”`m ) rSup { size 8{2} } =7 “.” “85” times “10” rSup { size 8{ – 3} } `m rSup { size 8{2} } } {}. Entering this value gives
Discussion
This is a practical average value, similar to the 120 V used in household power.
The emf calculated in [link] is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width ww size 12{w} {} and height ℓℓ size 12{l} {} in a uniform magnetic field, as illustrated in [link].
Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be emf=Bℓvemf=Bℓv size 12{“emf”=Bℓv} {}, where the velocity v is perpendicular to the magnetic field BB size 12{B} {}. Here the velocity is at an angle θθ size 12{θ} {} with BB size 12{B} {}, so that its component perpendicular to BB size 12{B} {} is vsinθvsinθ size 12{v”sin”θ} {} (see [link]). Thus in this case the emf induced on each side is emf=Bℓvsinθemf=Bℓvsinθ size 12{“emf”=Bℓv”sin”θ} {}, and they are in the same direction. The total emf around the loop is then
This expression is valid, but it does not give emf as a function of time. To