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.

Calculating the Emf Induced in a Generator Coil

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} {}:

emf=−NΔΦΔt.emf=−NΔΦΔt. size 12{“emf”= – N { {ΔΦ} over {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

ΔΦ=Δ(BAcosθ)=ABΔ(cosθ).ΔΦ=Δ(BAcosθ)=ABΔ(cosθ). size 12{ΔΦ=Δ ( ital “BA””cos”θ ) = ital “AB”Δ ( “cos”θ ) } {}

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

emf=NABΔt.emf=NABΔt. size 12{“emf”=N { { ital “AB”} over {Δt} } } {}

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

emf=200(7.85×10−3m2)(1.25 T)15.0×10−3 s=131V.emf=200(7.85×10−3m2)(1.25 T)15.0×10−3 s=131V. size 12{“emf”=”200” { { ( 7 “.” “85” times “10” rSup { size 8{ – 3} } ” m” rSup { size 8{2} } ) ( 1 “.” “25”” T” ) } over {“15” “.” 0 times “10” rSup { size 8{ – 3} } ” s”} } =”131″” V”} {}

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 find the time dependence of emf, we assume the coil rotates at a constant angular velocity ωω size 12{ω} {}. The angle θθ size 12{θ} {} is related to angular velocity by θ=ωtθ=ωt size 12{θ=ωt} {}, so that

Now, linear velocity
vv
is related to angular velocity
ωω by v=rωv=rω size 12{v=rω} {}. Here r=w/2r=w/2 size 12{r=w/2} {}, so that v=(w/2)ωv=(w/2)ω size 12{v= ( w/2 ) ω} {}, and

Noting that the area of the loop is A=ℓwA=ℓw size 12{A=ℓw} {}, and allowing for NN size 12{N} {} loops, we find that

is the emf induced in a generator coil of NN size 12{N} {} turns and area AA size 12{A} {} rotating at a constant angular velocity
ωω
in a uniform magnetic field BB size 12{B} {}. This can also be expressed as

where

is the maximum (peak) emf. Note that the frequency of the oscillation is f=ω/2πf=ω/2π size 12{f=ω/2π} {}, and the period is T=1/f=2π/ωT=1/f=2π/ω size 12{T=1/f=2π/ω} {}. [link] shows a graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal.

The fact that the peak emf, emf0=NABωemf0=NABω size 12{“emf” rSub { size 8{0} } = ital “NAB”ω} {}, makes good sense. The greater the number of coils, the larger their area, and the stronger the field, the greater the output voltage. It is interesting that the faster the generator is spun (greater ωω size 12{ω} {}), the greater the emf. This is noticeable on bicycle generators—at least the cheaper varieties. One of the authors as a juvenile found it amusing to ride his bicycle fast enough to burn out his lights, until he had to ride home lightless one dark night.

[link] shows a scheme by which a generator can be made to produce pulsed DC. More elaborate arrangements of multiple coils and split rings can produce smoother DC, although electronic rather than mechanical means are usually used to make ripple-free DC.

In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or the kinetic energy of wind. [link] shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator.

Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motor becomes a generator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In Back Emf, we shall further explore the action of a motor as a generator.