# Electromagnetism Exercise

Discussion in 'Electronics Homework Help' started by PostMortem, Jul 27, 2012.

1. ### PostMortem

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Jul 27, 2012
http://www.teiath.gr/userfiles/s.ath...im_2011_12.pdf

I have some trouble in exercise 1 and 2.

Exercise 1 says: There is a load q=-1μC in point A in axis x that is 0.15m form point O on the start of the axis. In the same time a second load Q is in point B of the axis Z that is 0.10m from point O on the start of the axis. The loads q and Q are moving with speeds u1 and u2. The direction of the two speed is shown on the graph, while we know that |u1|=10^7 m/s and |u2|=5*10^7 m/s. If the magnectic field on the point O is equal with zero calculate: a) The value and the sign for the load Q
and b) The electric field on point O

Exercise 2 says: Three loads Q1=-1nC, Q2=+2nC and Q3 are on axis x an specifically Q1 on the start of the axis, Q2 on the point X=0.1m and Q3 on the point X=0.3m

The loads Q1 an Q2 are moving with speeds u1 and u2, while the load Q3 is not moving. |u1|=|u2|=10^8 m/s.
If the total electric force that is being wielded on load Q1 equals zero calculate:
A) The value and the sign for load Q3
B)The magnetic force on load Q1.

2. ### john monks

693
2
Mar 9, 2012
Your pdf is not coming through.

16
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Jul 27, 2012
4. ### john monks

693
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Mar 9, 2012
I guess I don't understand the premise. For one thing the velocities look like they're getting close to relativistic velocities. But assuming they are not you know that the magnetic field generated by a moving charge is perpendicular to the direction of the moving charge. This has been discovered by experimentation. Now you know the polarity of the other charge and should be able to figure out the magnitude of the magnetic field. The formula has been discovered by experimentation and you should be able to look that up in your book.
The second problem can be figured by using similar logic.
Just remember this: use only physics and definitions to do your work and if you can't derive a formula from that don't use it.

Last edited: Jul 28, 2012
5. ### PostMortem

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Jul 27, 2012
I'm sorry but i don't understand how that tells me the polarity of the other charge...
I gotta say that I never really liked physics so this whole thing looks like jibberish to me :/
If you could direct me to the formula you are talking about it would be very helpful
Also for the second part of the exercise arent any standard formulas that i can use? In my notes i didn't find any.

Thank you for taking the time to answer me

6. ### john monks

693
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Mar 9, 2012
I would love to give you a formula but believe me even if I did it would be of no help. And you could not find a formula in your notes because the professor wisely did not give you one. But the professor should have given you the characteristics of charged particles in motion. And I believe the reason you never liked physics is because nobody explained it to you in a way you could understand it. Going to formulas is the last thing you want to do because you will not remember it or understand it.

So if you have a point O in space and a charged particle that is to the right and traveling down and another particle above it traveling to the right the resulting magnetic fields on point O will add up because of the right hand rule.
That is a magnetic field occurs perpendicular to the travel of a charged particle.
The magnetic field changes with the strength of the charged particle.
The magnetic field reverses with a reversal of the charged particle.
The magnetic field reverses if the charged particle reverses it's direction of travel.
The magnetic field is inversely proportional to the square of the distance to the charged particle.
The magnetic field from multiple charged particles add up.

Please don't become confused by the terms. When you have a charged particle passing in front of you to the right a magnetic field exists where you head is. The top of your head will be one pole and the bottom will be the opposite magnetic pole. If the charged particle goes from right to left the magnetic polse reverse. And if you change the electrical charge of the particle th pole changes again. Now if you have two charged particles traveling around your head in one direction the magnetic field generated by the particles and up. And if you have two charged particles circling around your head at the same direction at the same distance but one particle is 1 coulomb and the other is -1 coulomb the resultant magnetic field adds up to zero field.
The field strength is directly proportional to the charge and inversely proportional to the distance squared.

Be patient.
Everything you need to solve this problem is contained right here.
The mathematics is simple algebra and can be easily derived from this posting. And you must have had mathematics through calculus to be at this level.
If this is not clear get your physics book and look at the magnetic effect of a moving charged particle.
I will happily explain this in another way if you wish.
I would give you the formula but I already got in trouble for doing that for somebody else. And besides giving you the formula would not be helpful. Very few students have a problem with the math. Most have problems understanding the physical concepts. And in your case Understanding the physics of charged particles in motion.
And in the end you must must study physics by only looking at physical characteristics discovered by people and definitions.

I strongly recommend that you work with your classmates.
Work in your homes, in the school library, or any other comfortable setting.
Bring food with you. Physics takes time and is mostly a group effort.

Last edited: Jul 29, 2012
7. ### PostMortem

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Jul 27, 2012
So the sign for the charge Q is + because the other charge is - and you said that the magnetic field adds up so since its zero it has to be + for the charge Q?
Those things you said about the magnetic field although they are helpful in understanding physics don't really help me with this particular exercise much exept for the sign. I don't know how or where to use the velocities or how to calculate the Q charge Also i need you to tell me about the electric field too and the magnetic and electric force or maybe point me to a website where i can read about it..Thank you again for you time.

Edit: I found that the electric force is given by: Fe=ke*(q1*q2)/r^2, is there a formula for the magnetic force too? Also is the magnetic force the same with the magnetic field and the electric force the same with the electric field? If not what's the difference?

Last edited: Jul 29, 2012
8. ### john monks

693
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Mar 9, 2012
Yes, there is a formula for the magnetic field too.

In this case you do not need it. You already have enough information to logically calculate the charge on B.

The magnetic field is not the same as the electric field.
The magnetic force is not the same as the electric force.
The magnetic force occurs when The poles of two magnets line up.
The magnetic force occurs when a magnet is perpendicular to the path of a moving charged particle.
The electric field is the effect of a charged particle in space or a changing magnetic field.
The electric force is a mechanical force created by two charged particles in space.
The electric field pull particles together when the charges of the particles are the opposite.
The electric field pushes particles apart when the charges of the particles are the same.
The magnetic field pulls when opposite poles are pointed at each other.
The magnetic field pushes when the same poles are pointed at each other.

Nobody knows exactly what a magnetic field, electric field, or gravity is. We are looking at the effects of things in nature. This is the best we can do for now.

The magnetic field is inversely proportional to the square of the distance distance of a charged particle moving parallel to the distance. So you are looking at two relationships and the magnetism created by the two.
Charge A is moving clockwise around O at a known distance.
Charge B is moving clockwise around O at a known distance.
Both are subjecting point O to a net zero magnetics field.
Therefore you simply have to combine both relationships together that causes the final answer to come out to zero.

Your professor should have shown you the relationship of a single charged particle in motion and the resultant magnetic field at some distance away. Maybe you email your professor.
The important things are:
the resultant magnetic field is inversely proportional to the distanced squared of the charged particles.
The magnetic field is proportional to the velocity of the charge
or (B is proportional to QV/r^2) for perpendicular velocity.

You are doing well but by your questions I can easily see where you are confused. You like almost every other student is not confused by the mathematics but by the physics and definitions. Once you have a full grasp of the characteristics of magnets and charged particles your homework problems will become second nature.

Last edited: Aug 1, 2012
9. ### PostMortem

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Jul 27, 2012
What about the second exersice where there are 3 charges? How do i find the sign for the third charge? I think it's negative because as you said the magnetic field adds up so if charge 1 is -1nC and charge 2 is +2nC that leaves +1nC so for the magnetic field to be zero the other charge has to be -1nC? Is that correct?

10. ### john monks

693
2
Mar 9, 2012
It is negative because the charge on Q2 is positive.
The value is such that Q3/(.3^2) plus Q2/(.1^2) equals zero. These equations are derived from the definitions.
Remember the electric field must be zero for a charged particle to have zero electric force on it.
And the electric field from a charged particle is directly proportional to the charge divided by the distance from the particle squared.
The affects of charge particles is additive.
The magnetic force on Q1 can only be zero because the magnetic field is moving straight towards Q1, not at an angle.

Keep in mind that we are treating these problems without regard to relativity. That would change things slightly.
My only concern is that U1 and U2 are traveling at 1/3 the speed of light.

Last edited: Aug 22, 2012
11. ### PostMortem

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Jul 27, 2012
With the formula you gave me I calculated Q3 to be -18nC. Is that correct?
So from what you are saying I understand that in the first exercise with the two charges we wanted the magnetic field to be zero and on the second exercise with the three charges we want the electric field to be zero?
Also in the first exercise I calculated the Q charge to be -1nC, is that right?

12. ### john monks

693
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Mar 9, 2012
Q3 is -18nC.
I misread the velocity vectors therefore I will have to recalculate.
I got something different for exercise 1.

13. ### PostMortem

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Jul 27, 2012
What did you got for exercise 1? I recalculated and got +0,5μC

Last edited: Aug 26, 2012
14. ### john monks

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Mar 9, 2012
Sorry for being so late but I noticed some contradictions in my own thinking.
But I did not get the same answer you got. But I am asking that another physicist chime in, work out the problem and post whatever they get. And then I will have some questions to ask. The basic principles as I stated I believe are correct. I just want someone to come in behind me to verify that I am doing everything correctly. They match up with principles stated in physics books.

I hope you don't give up on me.

You can post how you got your answer and maybe I can find an error.

15. ### wingnut

242
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Aug 9, 2012
As far as I understand problem 1 is that the two moving charges are creating equal and opposite forces at O. And the forces are directly proportional to the charge and velocity and inversely proportional to the distance from O squared.

Therefore q x u1 /(.15 x .15) =- Q x u2 / (.1 x .1) where u1 = 10^7 and u2 = 5 x 10^7 and q = -1 x 10^-6 C

Plug in the values and one finds Q which using the right hand curling rule must be positive charge.

That I think is the answer to 1a)

1b) uses the formula E = kQ/r^2 where k = 9 x 10^9

Work E out for both charges at O and then vector add them since they are at right angles.

16. ### john monks

693
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Mar 9, 2012
Wingnut, I'm glad you came in. Hope you can bail me out.
Now I know the magnetic affect is inversely proportional to the square of the distance, call it R. K is a constant so we can leave it out. We are only concerned with the ratio of two separate moving charged particles. they must come out as equal and opposite.
q=-1uC
V1=10^7 meters per second
R1=.15meters
V2=5*10^7 meters per second
R2=.1meters
Q=?

So my formula comes out as Q*V2/(R2^2)=-q*V1/R1^2).
Is this correct so far?

After your response I have some more questions.

Last edited: Sep 9, 2012
17. ### wingnut

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Aug 9, 2012
That seems 100% correct John.

18. ### PostMortem

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Jul 27, 2012
So to recap. In the first exercise the two charges are moving so the charge in question is moving so you used q x u1 /X1^2 =- Q x u2 / X2^2. In the second exercise two of the charges are moving but the charge in question is still so we used Q3/(X3^2)+Q2/(X2^2)=0 ..So if I understand correctly when the charge in question is moving we say they are equal but when it's still the two charges together equal zero? On the second exercise why didn't we use the u2 in the formula?

Regarding the electric field in O what do you mean at "right angles"?

For the magnetic force on Q1 in the second exercise i thought of using this: B=(μ0/4π)*Q1*(u1*r1/R1^2) where r1= the angle between Q1 and B using the right hand rule and then inserting the result in : Fb=Q1*u1*B

Is that correct?

Last edited: Sep 14, 2012
19. ### wingnut

242
9
Aug 9, 2012
What did you calculate Q to be?

E=kQ/r^2 where r stands for u1 or u2

Work out E for both q and Q at O.

Electric fields go from + to -
q thus has an attractive or upwards field at O
Q which is a positive charge is pushing or has a repulsive (to the left) field at O.

These fields at O therefore are up and to the left.
We vector add these using trig functions of sin, cos and tan to find magnitude and direction of both fields combined.

20. ### PostMortem

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Jul 27, 2012
I calculated Q to be +1μC. I understood the 1st exercise my questions above are about the second exercise with the 3 charges...