1. Jul 26, 2014

### iScience

Electrodynamics (Lorentz Force & Voltage) [edited]

Two part inquiry

PART I

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

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)$\equiv$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?

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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 $\sqrt{2}$

help is much appreciated. ty all!

Last edited: Jul 26, 2014
2. Jul 26, 2014

### Matterwave

Notice that on the left hand side is $u\tan(\theta)$ and on the right hand side is $v$. So it must be that $u\tan(\theta)=v$. Look at figure 7.11 to see why this might be

3. Jul 26, 2014

### iScience

yeah i actually caught that right after i posted it. I'm actually now stuck with a different set of questions.

but thanks!!