# Flux and Induced EMF vs displacement graphs

• Taniaz
In summary, the conversation discussed the variation of magnetic flux and induced emf in a coil moving at a constant speed in a non-uniform magnetic field. The induced emf can be calculated using the equation E = (delta N*flux) / delta t, and its variation with x can be sketched by considering the slope of the flux vs. x graph. At point 3, there is an instant change in the emf. The minus sign in Faraday's law is not necessary as the direction of positive and negative can be chosen. The speed of the coil can be calculated by using the equation E = (delta N*flux) / delta x * (delta x / delta t).
Taniaz

## Homework Statement

A small coil of wire is situated in a non-uniform magnetic field. The coil consists of 40 turns of wire and moves with a constant speed in a straight line. The coil has displacement x from a fixed point P. The variation with x of the magnetic flux in the coil is shown in the first graph.
Sketch the variation with x of the e.m.f E induced in the coil for values of x from x=0 to x=6.0 cm.

## Homework Equations

E = (delta N*flux) / delta t

## The Attempt at a Solution

I know the slope of the flux vs. x graph is the emf so in the region where the flux vs. x graph has a negative yet constant slope, I drew a horizontal line in the negative region for the emf vs x graph. And the region where the flux vs x graph has a positive slope, I drew a horizontal line in the positive region for the emf vs x graph.

I'm confused as to what happens at point 3? Will the emf vs x graph be vertical as I drew it (second graph)?
Thank you.

#### Attachments

• 17274762_10155085213328838_1025659159_n.jpg
69.3 KB · Views: 1,042
• 17238775_10155085213228838_1074607773_n.jpg
36.5 KB · Views: 1,128
you are essentially correct with your analysis. Strictly speaking the induced emf is given by -d(NΦ)/dt so perhaps your lines should be inverted.
Do you realize that the lines shown have different gradients?
At point 3 the change from one to the other is instantaneous...not much more to say about that !

Yes that makes sense. Yes they do have different gradients (shown by their steepness, that's why the second one was a bit higher than the first one).
So there should be a vertical line at 3 joining the two together?

Should we take the minus into account? Because sometimes they just write Faraday's law without the minus sign to indicate the magnitude only?

Taniaz said:
So there should be a vertical line at 3 joining the two together?
Yes.
How are you calculating the magnitude of the emf? What is the speed of the coil?
Taniaz said:
Should we take the minus into account? Because sometimes they just write Faraday's law without the minus sign to indicate the magnitude only?
Minus sign is not necessary, as it is up to you to choose which direction is plus or minus for induced emf.

In the first part of the question they told us that "The coil is moved at constant speed between point P and the point where x = 3.0 cm" and then asked us to calculate the flux linkage which I did by multiplying the number of turns of the coil by the change in the flux from x=0 till x=3.0 cm. In the next part of the question they gave us the induced emf and asked for the speed so I did E = delta N(flux) / delta t but this can be written as E = (delta N(flux) / delta x) ( delta x / delta t) and we can solve for delta x / delta t which is the speed.

Taniaz said:
"The coil is moved at constant speed between point P and the point where x = 3.0 cm" and then asked us to calculate the flux linkage which I did by multiplying the number of turns of the coil by the change in the flux from x=0 till x=3.0 cm.
Why did you calculate the change in flux? Is it specifically asked? Because "flux linkage" of the coil can be computed at any single value of x. You can compute it at x=0 and at x=6. Why take the difference of them?
Taniaz said:
In the next part of the question they gave us the induced emf and asked for the speed so I did E = delta N(flux) / delta t but this can be written as E = (delta N(flux) / delta x) ( delta x / delta t) and we can solve for delta x / delta t which is the speed.
Correct.

They asked for it from x=0 till x=3

Thank you

Taniaz said:
I know the slope of the flux vs. x graph is the emf
Shouldn't the slope of the flux vs. x graph be emf * velocity

V = (dN*flux/dt) = (dN*flux / dx) * (dx / dt)
dx / dt = velocity or speed
=> ( V / velocity ) = ( dN*flux / dx)

Last edited:

## 1. What is flux and induced EMF?

Flux is the measure of the flow of a physical quantity through a given surface. In the context of electromagnetism, it refers to the flow of electric or magnetic fields through a surface. Induced EMF, or electromotive force, is the voltage generated in a conductor when it is exposed to a changing magnetic field.

## 2. How are flux and induced EMF related?

According to Faraday's Law of Induction, a changing magnetic field will induce an EMF in a conductor. This means that as the flux through a surface changes, an induced EMF will be generated in any conductor within that surface.

## 3. What is a displacement graph?

A displacement graph is a plot of the displacement of an object over time. In the context of flux and induced EMF, it refers to a graph of the induced EMF generated in a conductor as a function of time.

## 4. How do displacement graphs relate to flux and induced EMF?

The displacement graph of a conductor will show a peak or spike whenever there is a rapid change in the flux through the conductor. This is because the induced EMF is directly proportional to the rate of change of flux. Therefore, the shape and amplitude of the displacement graph can provide information about the flux and induced EMF in a system.

## 5. What are some practical applications of flux and induced EMF vs displacement graphs?

Flux and induced EMF vs displacement graphs are commonly used in the design and analysis of electrical motors and generators. They can also be used to study the behavior of inductors and transformers. Additionally, these graphs are useful in measuring the efficiency and performance of various electrical systems.

• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
41
Views
4K
• Introductory Physics Homework Help
Replies
7
Views
458
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
330
• Introductory Physics Homework Help
Replies
35
Views
3K
• Introductory Physics Homework Help
Replies
11
Views
977
• Introductory Physics Homework Help
Replies
2
Views
2K