# Electromagnetic Induction and an electric generator

1. Mar 1, 2012

### physgrl

1. The problem statement, all variables and given/known data
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

2. Relevant equations

ε=-NΔ∅B/Δt

B=BAcosβ
3. 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 dont see what I am doing wrong? Thanks!

2. Mar 1, 2012

### MisterX

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: Mar 1, 2012
3. Mar 1, 2012

### physgrl

I dont understand. The greatest change is from 0T to 2.5T so woudnt that cause the greatest change in ∅B hence the greater emf?

4. Mar 1, 2012

disregard

5. Mar 1, 2012

### I like Serena

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: Mar 1, 2012
6. Mar 1, 2012

### physgrl

The angular velocity is given by 100rpm.

7. Mar 1, 2012

### I like Serena

Can you turn that into radians per second?

8. Mar 1, 2012

9. Mar 1, 2012

### I like Serena

Good!

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. Mar 1, 2012

### physgrl

its BA*-sin(β(t))*β(t)'
but what is β(t)?

is it 10.5*t?

11. Mar 1, 2012

### I like Serena

Yes.

An angle is equal to the (constant) angular velocity multiplied with the time.

12. Mar 1, 2012

### physgrl

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. Mar 1, 2012

### I like Serena

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. Mar 1, 2012

### physgrl

sin(theta)=1 is the max so is the max emf supposed to be=2.5T*(.03*.05)*10.5??

15. Mar 1, 2012

### I like Serena

Yep. :)

16. Mar 1, 2012

### physgrl

ohh wait *500

17. Mar 1, 2012

### I like Serena

Oh yeah, right, very good.
I'm sorry. I totally forgot about that.

18. Mar 1, 2012

### physgrl

ok! thanks for the help!

19. Mar 1, 2012

### I like Serena

You appear to do a lot with electromagnetism.
What are you into?

20. Mar 1, 2012

### physgrl

Im taking AP Physics right now but I wanna major in Electrical Eng. :)