How Does Magnet Position Affect EMF in a Coiled Wire System?

[timetraveller123]

Homework Statement

A bar magnet is attached to the end of a non-magnetic spring, with spring constant 484−1, as shown in the diagram. The other end of the spring is attached to a fixed force sensor which was zeroed when the magnet was at its equilibrium position, before being zeroed, the force sensor shows 4.8N. The magnet is right at the center of a coil of wires when the magnet is in its equilibrium position. Assume the spring has negligible mass and that friction is negligible. The coil has length _c and the magnet is has length _m ≈ _c/4. The spring is stretched by ₀ = 1 cm < and released at time = 0 such that it performs simple harmonic motion in the coil. When the magnet was initially moved from far above the coil to inside the center of the coil a plot of the EMF vs time was obtained. The area under this curve was found to be -0.003Vs.
1i)sketech magnetic flux through coil against distance x
ii)sketch slope of the above graph against x

2i)sketch force against time
ii)velocity of magnet dx/dt against time
iii)displacement of magnet x against time

3) sketch a well labelled graph of emf against time show how the various features of emf curve correspond to features on the force curve

Homework Equations

emf = - dΦ/dt
f = -kx

The Attempt at a Solution

for part 1i)
i don't quite get why would magnetic flux change with distance x what is making it very complicated for me is the fact that the length of the coil is longer than the magnet and this is nothing like a solenoid where the field line only diverge after leaving the solenoid for the most part
so any help is appreciated
[/B]

 

[andrevdh]

The field lines of the bar magnet curves around from its one pole to the other, north to south, and continues right throught the magnet.
The field lines spreads out as they exit the pole and continues outwards.

The induced emf that were measured is the result of sum total of the flux change through each of the coils of the magnet - each coil acts like a little cell and all the little emf's add up to produce the final emf measured.

This means that as the magnet oscillates up and down the flux changes through the individual coils since more field lines enter them when the pole is closer and less when its further away.

In the picture the coil is moving towards the stationary magnet:

 

[timetraveller123]

okay i see what you are saying but i am still not getting what the sketch should look like because if you shift the magnet up by a bit the magnetic flux leaving the coil through the top will increase but however more magnetic flux will enter the coil through the sides so how to do please help

 

[andrevdh]

I think they want you to sketch the shape of the graph as the magnet oscillates up and down.
This should be a function of the speed of movement and the linkage of the field lines.

 

[timetraveller123]

i am sorry sir i don't get how is the shape related to speed as that would be plotting phi against time i think
this is something i thinked of
d phi/dx * dx / dt = d phi/dt = emf
i really am not getting anywhere please help thanks

 

[TSny]

d phi/dx * dx / dt = d phi/dt = emf

Yes, this should be a useful relation for this problem.

In your first post, there are some typographical errors. The value of the spring constant looks odd. The symbols for the lengths of the magnet and coil are not clear. Same for the initial position of the magnet.

Are you given numerical values for the lengths of the magnet and coil?

Anyway, as a start, you should be able to make a rough sketch of the graph of the flux Φ through the coil as a function of the position, x, of the magnet. From that, you can make a rough sketch of dΦ/dx as a function of x.

 

[timetraveller123]

@TSny
i am sorry about the typo errors got messed up when i copied over really sorry
there is no numerical values for this question except for the amplitude of oscillations

Anyway, as a start, you should be able to make a rough sketch of the graph of the flux Φ through the coil as a function of the position, x, of the magnet.

that is exactly what i don't get please help with that part thanks for your interest and help

 

[TSny]

What can you say about Φ for very large values of |x|?

Think about starting the magnet far away on the negative x-axis and moving the magnet along the x-axis so that it passes through the coil and moves far away on the positive x-axis. Think about the flux at the beginning and end of this movement. Think about how the flux changes as you move the magnet through the coil. Should the function Φ(x) be an even function, an odd function, or neither?

 

[TSny]

Also, is there any way you can post the diagram for the problem? There might be some useful information there.

 

[timetraveller123]

but in this case the magnet stays inside and x is very small here

 

[timetraveller123]

ok sure

 

[timetraveller123]

can i give you the link cause when i try to post it gets blurred out

 

[timetraveller123]

http://lms.asknlearn.com/EdulearnNE...a314-1150b55dd504/SJPO_Special_Round_2016.pdf

 

[timetraveller123]

i know gving links is not allowed but i don't know how to get the photo here from the pdf sorry moderator please understand thanks

 

[TSny]

but in this case the magnet stays inside and x is very small here

For part 1 of the question, it does not say to plot Φ vs x for just small values of x. Anyway, 1 cm is not necessarily a very small distance. If the length of the coil is 0.5 cm, then 1 cm is not small relative to the size of the coil.

 

[TSny]

http://lms.asknlearn.com/EdulearnNE...a314-1150b55dd504/SJPO_Special_Round_2016.pdf

This link doesn't work for me.

 

[timetraveller123]

sorry the question i gave is not complete it is
1i)sketech magnetic flux through coil against distance x where x is the distance from the center of mass of magnet to the equilibrium position

ok but anyways does the sketch look like a binomial distribution what i mean is some thing like

 

[TSny]

Yes, that looks OK to me for the general shape. Φ should approach zero for |x| →∞. (It's probably not an actual binomial distribution or "gaussian".)

From some of the information given in the problem, I think you can get the value of Φ at x = 0.

 

[timetraveller123]

sorry for the late reply it at x =0 Φ = - 0.003 vs

 

[TSny]

sorry for the late reply it at x =0 Φ = - 0.003 vs

Yes, except for the negative sign. Your graph shows positive flux at x = 0. Note that the units Vs is equivalent to Tm² (or Webers, Wb).

Can you sketch the graph of dΦ/dx?

 

[timetraveller123]

Yes, except for the negative sign.

but then why is it given negative in the question

let me start all over again

is this correct is there any other numericals that i can insert if this is correct
for the next part qn 2)
i)
a = w²x
a = w²cos(wt)
ma = mw²Acoswt = f

ii)velocity of magnet = Awsin(wt)

iii) x= A sinwt

are these correct i will later post the sketches of these once you approve of my qn1)
thanks for your continued interest in this question

 

[TSny]

but then why is it given negative in the question

Faraday's law is often written with a negative sign: ε = -dΦ/dt. So, Φ is the negative of the integral of ε with respect to time.

let me start all over again
View attachment 206476
View attachment 206477
is this correct is there any other numericals that i can insert if this is correct

Yes, I think the graph would be approximately a straight line for small x. I don't see how to get any other numerical data for this graph. It would be nice if there were a way to get the slope of this graph, but I don't see any information in the problem that allows you to do this.

 

[timetraveller123]

sorry for the late reply
and by the way the spring constant is 484nm^-1

$2i)\\ f= - kx\\ f = -k Acos(wt)$

$qn2ii)\\ v = - Awsinwt\\ v_{max} = Aw = 0.01\sqrt{\frac{k}{m}} = 1m/s$

$x = Acos wt$

just realising i could also have put in the time value for one whole period which is 0.065 seconds is there any other numerical values necessary and is the sketches correct thanks

 

[TSny]

The shapes of the graphs look good. In calculating v_max I don't think you calculated √(k/m) correctly. What value did you use for the mass? Also, your value for the period of the motion does not look correct..

 

[timetraveller123]

sorry for the very very late reply had a lot of work sorry

so k is 484
and mass is 4.8/10 = 0.48kg
v_max = 0.317m/s

period is 0.1978 seconds
correct?

 

[TSny]

Yes, the period is about 0.20 s.

 

[timetraveller123]

what about my v max is it correct

 

[TSny]

what about my v max is it correct

That looks correct also.

 

[timetraveller123]

sorry for the late reply here is my next sketch

is it correct i have omitted markings of time

 

[TSny]

Can you label the tick marks of the time axis of the emf graph with numerical values? I don't think the force values are correctly positioned on the emf graph.

Did you use your relation dΦ/dt = dΦ/dx dx/dt? What is the time dependence of each factor on the right side?

 

[timetraveller123]

sorry for the very late reply my exams are starting soon and i have to focus on my academics hope you understand

i am having some problems

so using the relation

dΦ/dt = dΦ/dx *dx/dt
dx/dt = v
i decided to express v of x
so it is
v = w√(A² - x^2)
so the graphs looked like this

hence the final dΦ/dt turned to look like this

in other words a damped oscillation
but i am not understanding this
there is a certain level of symetry in the problem in that after every oscillation the emf should be the same at the same position as the before oscillation but in this sketch it seems to suggest that the emf would never reach the same value again as time progresses

 

[TSny]

I'm not following what you did here. You have $\varepsilon = - \frac{d \Phi}{dt} = -\frac{d \Phi}{dx} \frac{dx}{dt}$.

From post #21 you have that $\frac{d \Phi}{dx}$ is linear in $x$ with a negative slope (for small oscillations). So, it can be written $\frac{d \Phi}{dx} = -b x$, where $b$ is some positive number. I don't see a way to determine a value for $b$ from the data given in the problem.

So, $\varepsilon = -\frac{d \Phi}{dx} \frac{dx}{dt} = bxv$. Or, as a function of time, you have $\varepsilon (t)= bx(t)v(t)$. You already have expressions for $v(t)$ and $x(t)$.

 

[timetraveller123]

oh sorry i used phi(x) instead of dphi/dx i will try again
thanks for the fast reply

 

[timetraveller123]

ok since v(t) is cosine and x(t) is -sine they both have same frequencies
v*x = -1/2sin 2w scaled uo by a factor of b

 

[TSny]

Yes. The frequency of the emf is twice the frequency of the oscillation of the magnet.