# 23.4 Electric generators

 Page 1 / 6
• 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 $\theta =0º$ to $\theta =\text{90º}$ ) 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? 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 $\Delta t$ :

$\text{emf}=-N\frac{\Delta \Phi }{\Delta t}\text{.}$

We know that $N=\text{200}$ and $\Delta t=\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{ms}$ , and so we must determine the change in flux $\Delta \Phi$ to find emf.

Solution

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

$\Delta \Phi =\Delta \left(\text{BA}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)=\text{AB}\Delta \left(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)\text{.}$

Now, $\Delta \left(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)=-1\text{.}0$ , since it was given that $\theta$ goes from $\text{0º}$ to $\text{90º}$ . Thus $\Delta \Phi =-\text{AB}$ , and

$\text{emf}=N\frac{\text{AB}}{\Delta t}.$

The area of the loop is $A={\mathrm{\pi r}}^{2}=\left(3\text{.}\text{14}\text{.}\text{.}\text{.}\right)\left(0\text{.}\text{0500}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}=7\text{.}\text{85}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ . Entering this value gives

$\text{emf}=\text{200}\frac{\left(7\text{.}\text{85}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left(1\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{T}\right)}{\text{15}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{131}\phantom{\rule{0.25em}{0ex}}\text{V.}$

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 $w$ and height $\ell$ in a uniform magnetic field, as illustrated in [link] . 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 $\text{emf}=\mathrm{B\ell v}$ , where the velocity v is perpendicular to the magnetic field $B$ . Here the velocity is at an angle $\theta$ with $B$ , so that its component perpendicular to $B$ is $v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ (see [link] ). Thus in this case the emf induced on each side is $\text{emf}=\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , and they are in the same direction. The total emf around the loop is then

$\text{emf}=2\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \text{.}$

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 $\omega$ . The angle $\theta$ is related to angular velocity by $\theta =\mathrm{\omega t}$ , so that

$\text{emf}=\text{2}\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\omega t}\text{.}$

Now, linear velocity $v$ is related to angular velocity $\omega$ by $v=\mathrm{r\omega }$ . Here $r=w/2$ , so that $v=\left(w/2\right)\omega$ , and

#### Questions & Answers

Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, College physics -- hlca 1104. OpenStax CNX. May 18, 2013 Download for free at http://legacy.cnx.org/content/col11525/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics -- hlca 1104' conversation and receive update notifications?   By By Rhodes     By 