<< Chapter < Page Chapter >> Page >
  • 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.

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

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele 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!
QuizOver.com Reply
Practice Key Terms 3

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?

Ask