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Electromagnetic Induction

by physgrl
Tags: electromagnetic, induction
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physgrl
#1
Mar1-12, 01:53 PM
P: 138
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!
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MisterX
#2
Mar1-12, 02:01 PM
P: 577
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.
physgrl
#3
Mar1-12, 02:14 PM
P: 138
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?

SHISHKABOB
#4
Mar1-12, 02:27 PM
P: 614
Electromagnetic Induction

disregard
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#5
Mar1-12, 02:42 PM
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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?
physgrl
#6
Mar1-12, 02:49 PM
P: 138
The angular velocity is given by 100rpm.
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#7
Mar1-12, 02:50 PM
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Quote Quote by physgrl View Post
The angular velocity is given by 100rpm.
Can you turn that into radians per second?
physgrl
#8
Mar1-12, 02:54 PM
P: 138
its 10.5rad/s
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#9
Mar1-12, 03:00 PM
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Quote Quote by physgrl View Post
its 10.5rad/s
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.
physgrl
#10
Mar1-12, 03:13 PM
P: 138
its BA*-sin(β(t))*β(t)'
but what is β(t)?

is it 10.5*t?
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#11
Mar1-12, 03:15 PM
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Quote Quote by physgrl View Post
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.
physgrl
#12
Mar1-12, 03:26 PM
P: 138
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?
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#13
Mar1-12, 03:28 PM
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Quote Quote by physgrl View Post
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?
physgrl
#14
Mar1-12, 03:39 PM
P: 138
oh yes my bad!
sin(theta)=1 is the max so is the max emf supposed to be=2.5T*(.03*.05)*10.5??
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#15
Mar1-12, 03:40 PM
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Quote Quote by physgrl View Post
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. :)
physgrl
#16
Mar1-12, 03:41 PM
P: 138
ohh wait *500
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#17
Mar1-12, 03:41 PM
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Quote Quote by physgrl View Post
ohh wait *500
Oh yeah, right, very good.
I'm sorry. I totally forgot about that.
physgrl
#18
Mar1-12, 03:43 PM
P: 138
ok! thanks for the help!


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