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  • 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.

Calculating the emf induced in a generator coil

The generator coil shown in [link] is rotated through one-fourth of a revolution (from θ = to θ = 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?

The figure shows a schematic diagram of an electric generator. It consists of a rotating rectangular coil placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the coil. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this coil are connected to the two small rings. The two conducting carbon brushes are kept pressed separately on both the rings. The coil is attached to an axle with a handle at the other end. Outer ends of the two brushes are connected to the galvanometer. The axle is mechanically rotated from outside by an angle of ninety degree that is a one fourth revolution, to rotate the coil inside the magnetic field. A current is shown to flow in the coil in clockwise direction and the galvanometer shows a deflection to left.
When this generator coil is rotated through one-fourth of a revolution, the magnetic flux Φ changes from its maximum to zero, inducing an emf.

Strategy

We use Faraday’s law of induction to find the average emf induced over a time Δ t size 12{Δt} {} :

emf = N Δ Φ Δ t . size 12{"emf"= - N { {ΔΦ} over {Δt} } } {}

We know that N = 200 size 12{N="200"} {} and Δ t = 15 . 0 ms size 12{Δt="15" "." 0`"ms"} {} , and so we must determine the change in flux Δ Φ size 12{ΔΦ} {} to find emf.

Solution

Since the area of the loop and the magnetic field strength are constant, we see that

Δ Φ = Δ ( BA cos θ ) = AB Δ ( cos θ ) . size 12{ΔΦ=Δ \( ital "BA""cos"θ \) = ital "AB"Δ \( "cos"θ \) } {}

Now, Δ ( cos θ ) = 1 . 0 size 12{Δ \( "cos"θ \) = - 1 "." 0} {} , since it was given that θ goes from to 90º . Thus Δ Φ = AB size 12{ΔΦ= - ital "AB"} {} , and

emf = N AB Δ t . size 12{"emf"=N { { ital "AB"} over {Δt} } } {}

The area of the loop is A = πr 2 = ( 3 . 14 . . . ) ( 0 . 0500 m ) 2 = 7 . 85 × 10 3 m 2 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 3 m 2 ) ( 1 . 25 T ) 15 . 0 × 10 3 s = 131 V. 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.

Got questions? Get instant answers now!

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 w size 12{w} {} and height size 12{l} {} in a uniform magnetic field, as illustrated in [link] .

The figure shows a schematic diagram of an electric generator with a single rectangular coil. The rotating rectangular coil is placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the coil. The magnetic field B is shown pointing from the North to the South Pole. The North Pole is on the left and the South Pole is to the right and hence the direction of field is from left to right. The angular velocity of the coil is given as omega. The velocity vector v of the coil makes an angle theta with the direction of field.
A generator with a single rectangular coil rotated at constant angular velocity in a uniform magnetic field produces an emf that varies sinusoidally in time. Note the generator is similar to a motor, except the shaft is rotated to produce a current rather than the other way around.

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 = Bℓv size 12{"emf"=Bℓv} {} , where the velocity v is perpendicular to the magnetic field B size 12{B} {} . Here the velocity is at an angle θ size 12{θ} {} with B size 12{B} {} , so that its component perpendicular to B size 12{B} {} is v sin θ size 12{v"sin"θ} {} (see [link] ). Thus in this case the emf induced on each side is emf = Bℓv sin θ size 12{"emf"=Bℓv"sin"θ} {} , and they are in the same direction. The total emf around the loop is then

emf = 2 Bℓv sin θ . size 12{"emf"=2Bℓv"sin"θ} {}

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 ω size 12{ω} {} . The angle θ size 12{θ} {} is related to angular velocity by θ = ωt size 12{θ=ωt} {} , so that

emf = 2 Bℓv sin ωt . size 12{"emf"=Bℓv"sin"ωt} {}

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

Questions & Answers

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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