Gregg Wolfe; Erika Gasper; John Stoke; Julie Kretchman; David Anderson; Nathan Czuba; Sudhi Oberoi; Liza Pujji; Irina Lyublinskaya; Douglas Ingram

Learning Objectives

By the end of this section, you will be able to:

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.

Example 23.3

Calculating the Emf Induced in a Generator Coil

The generator coil shown in Figure 23.19 is rotated through one-fourth of a revolution (from $θ = 0º$ 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.

Figure 23.19 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 $Δ t$ :

emf = - N \frac{Δ Φ}{Δ t} .

23.11

We know that $N = 200$ and $Δ t = 15 . 0 ms$ , and so we must determine the change in flux $Δ Φ$ 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) .

23.12

Now, $Δ (cos θ) = - 1 . 0$ , since it was given that $θ$ goes from $0º$ to $90º$ . Thus $Δ Φ = - AB$ , and

emf = N \frac{AB}{Δ t} .

23.13

The area of the loop is $A = {πr}^{2} = (3.14 ...) (0.0500 m)^{2} = 7.85 \times 10^{- 3} m^{2}$ . Entering this value gives

emf = 200 \frac{(7.85 \times 10^{- 3} m^{2}) (1.25 T)}{15.0 \times 10^{- 3} s} = 131 V.

23.14

Discussion

This is a practical average value, similar to the 120 V used in household power.

The emf calculated in Example 23.3 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 $w$ and height $ℓ$ in a uniform magnetic field, as illustrated in Figure 23.20.

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.

Figure 23.20 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 $emf = Bℓv$ , where the velocity v is perpendicular to the magnetic field $B$ . Here the velocity is at an angle $θ$ with $B$ , so that its component perpendicular to $B$ is $v sin θ$ (see Figure 23.20). Thus in this case the emf induced on each side is $emf = Bℓv sin θ$ , and they are in the same direction. The total emf around the loop is then

emf = 2 Bℓv sin θ .

23.15

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 $ω$ . The angle $θ$ is related to angular velocity by $θ = ωt$ , so that

emf = 2 Bℓv sin ωt .

23.16

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

emf = 2 Bℓ \frac{w}{2} ω sin ωt = (ℓ w) Bω sin ωt .

23.17

Noting that the area of the loop is $A = ℓ w$ , and allowing for $N$ loops, we find that

emf = NAB ω sin ωt

23.18

is the emf induced in a generator coil of $N$ turns and area $A$ rotating at a constant angular velocity $ω$ in a uniform magnetic field $B$ . This can also be expressed as

emf = {emf}_{0} sin ωt,

23.19

where

{emf}_{0} = NAB ω

23.20

is the maximum (peak) emf. Note that the frequency of the oscillation is $f = ω / 2π$ , and the period is $T = 1 / f = 2π / ω$ . Figure 23.21 shows a graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal.

The first part of the figure shows a schematic diagram of a single coil electric generator. It consists of a rotating rectangular loop placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the loop. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this loop are connected to the two small rings. The two conducting carbon brushes are kept pressed separately on both the rings. The loop is rotated in the field with an angular velocity omega. Outer ends of the two brushes are connected to an electric bulb which is shown to glow brightly. The second part of the figure shows the graph for e m f generated E as a function of time t. The e m f is along the Y axis and the time t is along the X axis. The graph is a progressive sine wave with a time period T. The crest maxima are at E zero and trough minima are at negative E zero.

Figure 23.21 The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time.

{emf}_{0}

is the peak emf. The period is

T = 1 / f = 2π / ω

, where

f

is the frequency. Note that the script E stands for emf.

The fact that the peak emf, ${emf}_{0} = NAB ω$ , 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 $ω$ ), 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.

Figure 23.22 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.

The first part of the figure shows a schematic diagram of a single coil D C electric generator. It consists of a rotating rectangular loop placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the loop. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this loop are connected to the two sides of a split ring. The two conducting carbon brushes are kept pressed separately on both sides of the split rings. The loop is rotated in the field with an angular velocity w. Outer ends of the two brushes are connected to an electric bulb which is shown to glow brightly. The second part of the figure shows the graph for e m f generated as a function of time. The e m f is along the Y axis and the time t is along the X axis. The graph is a progressive and rectified sine wave with a time period T. The sine wave has only positive pulses. The crest maxima are at E zero.

Figure 23.22 Split rings, called commutators, produce a pulsed DC emf output in this configuration.

Example 23.4