Average induced EMF and average induced current of a flipped loop?

Click For Summary
The discussion focuses on calculating the average induced EMF and current in a wire loop flipped in a magnetic field. The average induced EMF can be determined using the change in magnetic flux, ΔΦ, divided by the time interval Δt, rather than considering individual positions of the loop. The average induced current is then calculated using the induced EMF divided by the resistance R. Participants clarify that the EMF arises from the rate of change of magnetic flux, not from separate EMF values at different positions. The conversation emphasizes understanding the relationship between magnetic flux and induced EMF for accurate calculations.
Violagirl
Messages
112
Reaction score
0

Homework Statement


A wire loop of resistance R and area A has its normal along the direction of a uniform magnetic field, B. The loop is then flipped over in a time Δt so that its normal is opposite to the field. a) Calculate the average induced EMF. b) Find the average induced current. c) If the field is out of the page, what is the direction of the induced current? d) What is the total charge transported through the circuit in the interval Δt?


Homework Equations


IB = ∫NBA sin θ

εind = dIB/ dt

θ = ωt


The Attempt at a Solution



I've drawn the situation out on paper. In the beginning, it seems that B and A are uniform to one another at angle of cos θ = 0 degrees, which equals 1.

After the loop is flipped, B and A are no longer uniform as A will be downwards with B in the same, original direction. They will be an angle of cos θ = 180 degrees, which equals -1.

I also know that since they are looking for the average induced EMF, I will need to use the equation:

Eind = ΔIB/Δt

So I will need to find the separate EMF values of each position of the loop.

I know then that for the average induced current, it will be similar in that I will need:

Iind = ΔEind/ΔR

From here though, I'm confused on if I'm starting it correctly. In the first situation, I thought that B and A were uniform to one another so I took:

Ib = ∫NBA sin ωt

Ib = -NBAω cos ωt

so then Ib = NBAω

But then taking the derivative to find Eind is where I'm having trouble. If I wait to find the derivative of -NBAω cos ωt, we get NBAω sin ωt, which give an answer of 0 since sin 0 degrees is equal to 0 and it doesn't seem right...
 

Attachments

  • Flipped loop.jpg
    Flipped loop.jpg
    14.6 KB · Views: 1,585
Physics news on Phys.org
Violagirl said:
Eind = ΔIB/Δt

So I will need to find the separate EMF values of each position of the loop.
There are not "separate EMF values of each position". There is a flux at each position. The EMF arises from the rate of change of flux.
Since you only want the average EMF, You don't need to worry about individual positions at all. ΔIB in your equation represents the total change of flux between the two positions. (To avoid confusion with current, I'll use ##\Phi## instead.) Since you only care about the average EMF over time Δt, you can just take the difference in flux between the two positions, ##\Delta\Phi##, and divide by Δt.
What is the flux through the loop in the initial position?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
1K
Replies
8
Views
2K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
8
Views
1K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 8 ·
Replies
8
Views
4K