A-level Physics - Faradays Law and Electromagnetic Induction

[I like number]

Homework Statement

The problem can be found http://www.drewett-gray.co.uk/alevel/ff&e/papers/ffe_04_Jun.pdf" (page 10 and 11) Question 4 part B i and ii
Mark Scheme if needed http://www.drewett-gray.co.uk/alevel/ff&e/papers/FFE_04_Jun_MS.pdf"

Homework Equations

Faradays law: The induced emf is directly proportional to the rate of change of flux linkage.
Induced EMF = flux change/time
Induced EMF = Blx/t = Blv

The Attempt at a Solution

Question 4 Part B)i
I have attempted this by following the method of a similar problem in my revision guide.
So after 0.2 seconds the coil enters the field...
Information I have:

N=1250
B=0.032 T
V=0.10 ms^-1

Induced EMF = Blv*N = 0.032*0.02*0.1*1250
=0.08 = 80mV as required
I've used this because (l*v = l * x/t which is the area cut per unit time). I know it got the right answer but I'm not sure if my method is legitimate or not.

Part B ii)
This part has been giving me real trouble.
The way I understand it there will be no reading on the voltmeter until 0.2 seconds because the coil has not entered the field.
I'm not sure what happens when it enters the field, but I think the voltage will increase as the coil enters the field, because more and more coil is cutting the field per unit time. Then once all the coil is in the field it will be constant for a while, and then will decrease as the coil leaves the field.
However the markscheme talks about having some graph in the negative region, and I'm not sure why this should be the case.
Another thing I was thinking was trying to work out the voltage at certain times to get a rough idea of the shape of the graph.


Thanks

I Like Number

 

[Stonebridge]

While the coil is entering the field, the induced voltage will remain constant at that value you calculated, 80mV.
This is because the flux linking the coil is increasing uniformly as it enters the field. More and more flux links as it moves. As Faraday's Law says the emf depends on the rate of change of flux, then so long as the rate change is uniform, and it is, the emf will be constant.
When fully in the field, there is no further change in the flux linking the coil so the induced emf will be zero.
When leaving the field, the rate of change is the same as when it entered, except the flux linkage is decreasing now, not increasing. So the emf induced will again be 80mV but in the opposite direction. (ie negative)
When fully out of the field, the emf will be zero again.

 

[I like number]

Thank you! That made a lot of sense and I understood it :D
I'm going to mull over it, have another shot at the question a little later and let you know how I get on/ if there is anything I don't fully grasp.

Thanks once again.

 

[Stonebridge]

Thank you! That made a lot of sense and I understood it :D
I'm going to mull over it, have another shot at the question a little later and let you know how I get on/ if there is anything I don't fully grasp.

Thanks once again.

You're welcome.

 

[rdl]

2

I would say that your method for part B)i is legitimate. You have correctly used Faraday's law and the formula for induced EMF to calculate the voltage. However, it is always important to double check your calculations and make sure you are using the correct units.

For part B)ii, your understanding is correct. As the coil enters the field, the voltage will increase due to more coil cutting the field per unit time. Once the entire coil is in the field, the voltage will remain constant until the coil starts to leave the field, at which point the voltage will decrease.

The reason for the negative region in the graph is because the coil is moving in and out of the field, which results in a changing magnetic flux. According to Lenz's law, the induced EMF will always oppose the change in flux. So when the coil is leaving the field, the induced EMF will be in the opposite direction compared to when it entered the field, resulting in a negative voltage.

To get a rough idea of the shape of the graph, you could calculate the voltage at different times as you suggested. This will give you a better understanding of how the voltage changes as the coil moves in and out of the field.