Obtaining potential from Lorentz Force

In summary, the conversation discusses the situation of pulling a wire through a B-field and trying to derive the work done on a unit charge in this scenario. The conversation also discusses a potential mistake in the approach of deriving the work and the four forces involved in the situation, highlighting the need to analyze and consider each force carefully. Ultimately, the key is to determine which forces contribute to the work done on the unit charge.
  • #1
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Two part inquiry


PART I


Situation: I have a B-field and I'm pulling a straight piece of wire through that B-field


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You pull the wire perpendicularly through the B-field at velocity v; charges between points a and b will experience the Lorentz Force and will begin to move along the wire with velocity u, and thus will experience a force opposite to the direction which you are pulling the wire.



Goal: Trying to derive the work done on a unit charge the classical way (by integrating along the path it takes)


Let us define the force per unit q to pull the wire through the field: f(pull)[itex]\equiv[/itex]uB


$$\int f_{pull} \cdot dl = (uB) \frac{y}{cosθ}sinθ= vBhy$$



Problem: The vector f(pull) should be pointing outward (opposite the direction of wire pulling). so then shouldn't there be a negative sign somewhere?

--------------------


PART II


I decided to try and derive it my own way and i ran into a problem somewhere but i can't seem to spot where it is.


I said that, there is a vertical work component and a horizontal work component.

vertical work component

(From the initial pull)

$$W_y= \int (q\vec{v} \times \vec{B}) \cdot dy = (qvB)y= qvBy$$


horizontal work component

$$W_x= \int (q\vec{u} \times \vec{B}) \cdot dx = quBx =quBytanθ$$


vectorial sum

(yes i know work is a scalar quantity. but there is a work in x direction and work in the y direction so i thought i'd try this approach)

$$W= \sqrt{(qvBy)^2+(quBytanθ)2}= qBy\sqrt{v^2+u^2tan^2θ}$$

but since utanθ=v

$$W=qBy\sqrt{2v^2}=\sqrt{2}qvBy$$

i'm off by a factor of [itex]\sqrt{2}[/itex]


Could someone please help me find the mistake in doing it this way?


help is much appreciated. ty all!
 
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  • #2
You are an observer at rest with the B field, watching the wire move by to the right.

There are 4 forces involved:
1. Lorentz force on the unit charge due to the wire motion in the x direction.
2. Lorentz force due to the motion of the charge in the y direction.
3. Since the force in 2 doesn't pull the charge out of the wire in the -x direction there must be a third force keeping the charge in the wire.
4. There is also a force along the wire associated with the collision of the electrons as they move up the wire.
This force obviously represents work since it's heating up the wire.

You need to analyze these four forces, determine which ones contribute to work done on the unit charge as it moves thru the wire and around the rest of the closed loop. Remember that the B field does no work on any charge. So, hint: think about the last two forces!
 

FAQ: Obtaining potential from Lorentz Force

What is the Lorentz Force?

The Lorentz Force is a fundamental concept in electromagnetism that describes the force exerted on a charged particle moving in an electromagnetic field.

What is potential in the context of Lorentz Force?

In the context of Lorentz Force, potential refers to the potential energy of a charged particle in an electromagnetic field. It represents the work required to move the particle from one point to another in the field.

How is potential obtained from the Lorentz Force?

Potential can be obtained from the Lorentz Force by using the formula V = -∫F⋅dr, where V is the potential, F is the Lorentz Force, and dr is the displacement of the particle. This integral is evaluated along the path of the particle in the electromagnetic field.

What are the units of potential in the Lorentz Force equation?

The units of potential in the Lorentz Force equation are Joules (J). This is because potential is a form of energy and energy is measured in Joules.

How is potential from Lorentz Force used in practical applications?

Potential obtained from the Lorentz Force is used in a variety of practical applications, such as in electric motors, generators, and particle accelerators. It is also used in the study of plasma physics and in the design of electronic devices.

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