Electric Field Direction for Electron Moving Parallel to Wire

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Homework Help Overview

The problem involves a long, straight wire carrying a current of 25.0 A, with an electron moving parallel to the wire at a velocity of 250 km/s, positioned 2.00 cm away from it. The discussion centers on determining the electron's initial acceleration and the required direction and magnitude of a uniform electric field to maintain its parallel trajectory with the wire.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the magnetic field and the Lorentz force acting on the electron. There are questions regarding the direction of the electric field and the use of the right-hand rule versus Fleming's left-hand rule to determine the force direction. Some participants express confusion about the direction of the electric field relative to the wire.

Discussion Status

Participants are actively engaging with the problem, checking calculations, and clarifying concepts related to the direction of forces and fields. There is recognition of mistakes in initial calculations, and some guidance has been provided regarding the use of rules for determining directions. Multiple interpretations of the direction of the electric field are being explored.

Contextual Notes

There is mention of using conventional current and confusion regarding the mass of the electron, which has led to discrepancies in calculations. Participants are also reflecting on their understanding of the rules for determining force directions in electromagnetic contexts.

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Homework Statement


4 A long, straight wire carries a current of 25.0 A. An electron is fired parallel to this wire with a velocity of 250 km.s-1 in the same direction as the current, 2.00 cm from the wire.
a) Find the magnitude and direction of the electron’s initial acceleration.
b) What should the magnitude and direction of a uniform electric field be that will allow the electron to continue to travel parallel to the wire?



Homework Equations





The Attempt at a Solution


B=kI/d
B=2.5*10-4 T
F=Bqv
F=1*10-17N
F=ma
a=1.099*1012m/s2

E=F/q
E=62.5N/c
I'm relatively confident with my numerical answers although I would appreciate someone looking over my basic procedure. I run into trouble tying to explain the direction of the field.
Were using conventional current (don't ask me why) and the best indication of the direction I can give is "at 900 away from the wire", thoughts?
Thanks.
 
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Check if you used proper value for the electron mass.

As for direction of electric current: By definition, the electric current flows in the direction of motion of the positive particles. When an electron is concerned, we speak about its velocity and the current it carries flows in the direction opposite to the velocity.

When you calculate the Lorentz force, it is [tex]\vec {F}=q \vec{v}\times \vec{b}[/tex]. You can clearly see the direction of F from the vector product, but there is a right-hand rule also for this. If v and B are perpendicular as is the case here, imagine a coordinate system with x-axis pointing in the direction of v, the y-axis pointing in the direction of B. The force points in the positive z direction if it is a positive charge. You have an electron , so the force will point in the negative z direction. Check if it is away or towards the wire. The electric force should be opposite, but the electric field is opposite again to the force as the charge of electron is negative.

ehild
 
Thanks ehild, yeah I did use the wrong value for mass, now I get a=6*〖10〗^9 m/s^2...
Regarding the direction should it be "90degrees towards the wire"? I always seem to get the exact opposite direction on these problems for some reason?
really appreciate you picking up on my stupid little mistakes- I would never have found that when checking.
 
pat666 said:
Thanks ehild, yeah I did use the wrong value for mass, now I get a=6*〖10〗^9 m/s^2...

It is even worse. The mass of electron is 9.1*10^-31 kg.
Regarding the direction should it be "90degrees towards the wire"? I always seem to get the exact opposite direction on these problems for some reason?

I think, yes, but who knows if there are so many opposite directions?

You seem to have the custom to contradict :smile:

ehild
 
Just realized when you said to check my mass I went ahead and googled mass of proton for some reason--brain explosion!
 
pat666 said:
Thanks ehild, yeah I did use the wrong value for mass, now I get a=6*〖10〗^9 m/s^2...
Regarding the direction should it be "90degrees towards the wire"? I always seem to get the exact opposite direction on these problems for some reason?
really appreciate you picking up on my stupid little mistakes- I would never have found that when checking.

I think you shoulb be using Fleming's left hand rule to find the direction of the electron,
it should be away from the wire. Or you can try to find out the vector multipler of IxB by certain
right hand ehilid mentioned. (two are different)
 
pat666 said:
Just realized when you said to check my mass I went ahead and googled mass of proton for some reason--brain explosion!


Never mind...:smile:

ehild
 
Terocamo said:
I think you shoulb be using Fleming's left hand rule to find the direction of the electron,
it should be away from the wire. Or you can try to find out the vector multipler of IxB by certain
right hand ehilid mentioned. (two are different)
I got away from wire by using Flemings LHR to but I am terrible at finding directions for problems like this. so is it toward or away from the wire?
 
pat666 said:
I got away from wire by using Flemings LHR to but I am terrible at finding directions for problems like this. so is it toward or away from the wire?

I'd say LHR is easier to use but [tex]\vec{I}[/tex] x [tex]\vec{B}[/tex] (by fanning your palm from I to B, your tumb is the direction of force) is faster.
It is away from the wire if you are taking about direction of electron movement.
 
  • #10
I tried to make a drawing to show the magnetic force on that electron. But the direction of the force can also be obtained if you think of two parallel current-carrying wires: If the current flows in the same direction in the wires, they attract each other. In case of opposite direction of currents, the wires repel each other.
The electrion moves in the same direction as the current flows in the wire, but it is a negative charge, so it represents an opposite current.
Think it over to be sure of your answer about the direction of the electric field.


ehild
 

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