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Electromagnetic Induction 
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#1
Mar112, 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 Δt_{half 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
Mar112, 02:01 PM

P: 581

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. 


#3
Mar112, 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?



#4
Mar112, 02:27 PM

P: 614

Electromagnetic Induction
disregard



#5
Mar112, 02:42 PM

HW Helper
P: 6,187

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? 


#6
Mar112, 02:49 PM

P: 138

The angular velocity is given by 100rpm.



#8
Mar112, 02:54 PM

P: 138

its 10.5rad/s



#9
Mar112, 03:00 PM

HW Helper
P: 6,187

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
Mar112, 03:13 PM

P: 138

its BA*sin(β(t))*β(t)'
but what is β(t)? is it 10.5*t? 


#11
Mar112, 03:15 PM

HW Helper
P: 6,187

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


#12
Mar112, 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?



#13
Mar112, 03:28 PM

HW Helper
P: 6,187

What are the maximum and minimum values the sine can take? 


#14
Mar112, 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?? 


#16
Mar112, 03:41 PM

P: 138

ohh wait *500



#17
Mar112, 03:41 PM

HW Helper
P: 6,187

I'm sorry. I totally forgot about that. 


#18
Mar112, 03:43 PM

P: 138

ok! thanks for the help!



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