Electromagnetic Induction and an electric generator

In summary, the conversation discusses the calculation of the maximum emf produced by an electric generator consisting of a rectangular loop of wire in a uniform magnetic field. The maximum emf is found by taking the derivative of the flux through the loop with respect to time, using the chain rule to account for the changing angle of the loop. The maximum emf is equal to 2.5T*(.03m*.05m)*10.5rad/s*500, or approximately 19.6 V.
  • #1
physgrl
138
0

Homework Statement


An electric generator consists of n = 500 turns of wire formed into a rectangular loop with a length of 5 cm and width of 3 cm placed in a uniform magnetic field of 2.50 T. What is the maximum value of the emf produced when the loop is spun at f = 100 rpm about an axis perpendicular to B?

a. 19.6 V
b. 144 V
c. 95.3 V
d. 79.2 V
e. 60.3 V

Homework Equations



ε=-NΔ∅B/Δt

B=BAcosβ

The Attempt at a Solution



For a maximum emf there must be a maximum change in either B or A, but because A is constant B must be the variable changing. As the loop rotates B varies from 0T to 2.5T and it takes half a loop to do that.

ε=NAB/Δt
ε=500*(.03m*.05m)*(2.5T)/Δt

Δthalf a revolution:
=>100rpm*1min/60s=1.67rev/sec
=> 1 revolution takes .6s
=> .5 revolutions take .3s

ε=500*(.03m*.05m)*(2.5T)/Δt
ε=500*(.03m*.05m)*(2.5T)/.3s
ε=6.25V

The answer I get is not in the options, and I don't see what I am doing wrong? Thanks!
 
Physics news on Phys.org
  • #2
The emf, ε, is related to the instantaneous rate of change in the flux, d∅B/dt. You used the value of the constant magnetic field times the area for this, which is incorrect.

The value you used, (.03m*.05m)*(2.5T), is the maximum flux though the coil. Instead this should be a maximum magnitude* rate of change of this flux (a value of d∅B/dt at some time).

You could start by expressing ∅B as a function of time.*I would pick the most negative value of d∅B/dt, that way ε takes on the highest positive value.
 
Last edited:
  • #3
I don't understand. The greatest change is from 0T to 2.5T so woudnt that cause the greatest change in ∅B hence the greater emf?
 
  • #4
disregard
 
  • #5
Hi physgrl!

The correct form is ##\mathcal{E}=-N{d\Phi \over dt}##.
where ##{d\Phi \over dt}## is the derivative of ##\Phi## with respect to t.You already know that ##\Phi=B A \cos \beta##.
And actually B and A are both constant, but ##\beta## is not.

##\beta## is the angle of the loop with the magnetic field.
Its derivative is the angular velocity.
Do you know how to find the angular velocity from the frequency with which the loop is spun?Furthermore, do you know how to take the derivative of ##\Phi = BA\cos \beta## with respect to t?
 
Last edited:
  • #6
The angular velocity is given by 100rpm.
 
  • #7
physgrl said:
The angular velocity is given by 100rpm.

Can you turn that into radians per second?
 
  • #8
its 10.5rad/s
 
  • #9
physgrl said:
its 10.5rad/s

Good! :smile:

Now you need the derivative of ##BA\cos(\beta(t))##.
For that you will need to apply the chain rule of differentiation.
Do you know how that works, or is that outside of the scope of your class material?
If it is, we may need to find another method to find the result.
 
  • #10
its BA*-sin(β(t))*β(t)'
but what is β(t)?

is it 10.5*t?
 
  • #11
physgrl said:
its BA*-sin(β(t))*β(t)'
but what is β(t)?

is it 10.5*t?

Yes.

An angle is equal to the (constant) angular velocity multiplied with the time.
 
  • #12
So the derivative is 2.5T*(.03*.05)*sin(10.5t)*10.5t but what is the time I should plug in? how do i know what makes the max emf?
 
  • #13
physgrl said:
So the derivative is 2.5T*(.03*.05)*sin(10.5t)*10.5t but what is the time I should plug in? how do i know what makes the max emf?

You should have a derivative of 2.5T*(.03*.05)*-sin(10.5t)*10.5.

What are the maximum and minimum values the sine can take?
 
  • #14
oh yes my bad!
sin(theta)=1 is the max so is the max emf supposed to be=2.5T*(.03*.05)*10.5??
 
  • #15
physgrl said:
oh yes my bad!
sin(theta)=1 is the max so is the max emf supposed to be=2.5T*(.03*.05)*10.5??

Yep. :)
 
  • #16
ohh wait *500
 
  • #17
physgrl said:
ohh wait *500

Oh yeah, right, very good.
I'm sorry. I totally forgot about that. :blushing:
 
  • #18
ok! thanks for the help!
 
  • #19
You appear to do a lot with electromagnetism.
What are you into?
 
  • #20
Im taking AP Physics right now but I want to major in Electrical Eng. :)
 
  • #21
physgrl said:
Im taking AP Physics right now but I want to major in Electrical Eng. :)

Good for you! :approve:
See you next time.
 
  • #22
:D ok!
 

What is electromagnetic induction?

Electromagnetic induction is the process by which a changing magnetic field can induce an electric current in a conductor. This was first discovered by Michael Faraday in the 1830s.

How does an electric generator work?

An electric generator works by using a rotating magnet to create a changing magnetic field, which then induces an electric current in a coil of wire. This current can then be harnessed for various purposes, such as powering electrical devices or generating electricity for homes and businesses.

What is the difference between AC and DC generators?

AC (alternating current) generators produce an electric current that periodically changes direction, while DC (direct current) generators produce a steady current that only flows in one direction. This is due to the different ways in which the magnetic field is rotated in each type of generator.

What is the relationship between electricity and magnetism in electromagnetic induction?

Electricity and magnetism are closely related in electromagnetic induction. A changing magnetic field can produce an electric current, and an electric current can produce a magnetic field. This phenomenon is known as electromagnetism and is a fundamental principle in understanding how electricity and magnetism interact.

What are some real-life applications of electromagnetic induction and electric generators?

Electromagnetic induction and electric generators have many practical applications, including producing electricity for everyday use, powering electric motors in vehicles and machinery, and generating energy from renewable sources such as wind and hydro power. They are also used in devices like transformers, generators, and motors.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
12K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
8
Views
2K
Back
Top