# Line Integral for Electromagnetic Force

• gibberingmouther
In summary, the problem is that the equation for F can be simplified if the x and y components of the vector r are rewritten as negative numbers. However, this change makes the equation for F negative in a way that is not obvious. The author's solution is to rewrite the equation for F using the unit vector ##\hat r## which points in the opposite direction.

#### gibberingmouther

http://web.mit.edu/sahughes/www/8.022/lec01.pdf

So I'm trying to understand how to get from F = ∫[(Q*λ)*dL*r]/(r^2) to F=∫q*λ*[(xx+ay)/(a^2+x^2)^(3/2)]*dx

Like I don't understand why the x and y components of r are negative, or why "The horizontal r component is obviously zero: for every element on the right of the midpoint, there is an element on the left whose force magnitude is equal, but whose horizontal component points in the opposite direction."

This was kind of a starting point for trying to understand some of the more "fundamental" Electricity and Magnetism equations and ideas, but I got stuck.

I cannot read your formulae well. You should also better explain the physics context the integrals are needed for. Maybe then we'll be able to help you with your problem.

gibberingmouther said:
I don't understand why the x and y components of r are negative
I assume you're referring to the unit vector ##\hat r## which he writes near the bottom of page 7: $$\hat r = -\hat x \cos \theta - \hat y \sin \theta$$ According to the diagram in the middle of the page, ##\hat r## starts at the charge segment dx and points downwards and to the left. His coordinate system is set up as xy systems usually are: x is positive to the right and negative to the left, y is positive upwards and negative downwards. This determines the signs on the components of ##\hat r##.

(Tip: to learn how to write equations "properly", see here: https://www.physicsforums.com/help/latexhelp/)

or why "The horizontal r component is obviously zero: for every element on the right of the midpoint, there is an element on the left whose force magnitude is equal, but whose horizontal component points in the opposite direction."

Draw a new version of the diagram in the middle of page 7, with a second charge segment dx located the same distance x from the midpoint, but on the left side. Draw the unit vector ##\hat r## associated with that second dx. Which way does it point, what are the signs on its x and y components, and how do they compare to the other dx? What happens when you add the contributions from the two dx's, as part of the integral on the next page?

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gibberingmouther said:
So I'm trying to understand how to get from F = ∫[(Q*λ)*dL*r]/(r^2) to F=∫q*λ*[(xx+ay)/(a^2+x^2)^(3/2)]*dx
I cannot read your equations. Could you write them in LaTex and also explain what the variables are? What are x and L, what are Q and q, etc.

I figured out how to use Latex. I originally thought I had to use a separate program to format the Latex code but then I re-read the Latex Primer here on Physics Forums and realized that it is much easier than that!

So the math I was stuck on is:

##\vec F = \int (Q \times λ \times dl \times r)/(r^2)## somehow becomes ##\vec F = - \int_{-L/2}^{L/2} q \times λ \times [(x \times \hat x +a \times \hat y)/(a^2+x^2)^{3/2}] dx##

We have a rod of length L horizontally there in space with a charge Q smeared uniformly on it and a test charge q somewhere else with relation to it.

As for the variables
* ##\lambda## = Q / L = the line charge density
* Q is the value of the charge that is uniformly smeared on the rod
* q = value of the test charge in coulombs
* x is the horizontal axis and the rod smeared with charge we're talking about goes from -L/2 to L/2 where L is the length of the horizontal rod
* ##\hat r## is the length from the test charge to a point on the rod and ##\hat x## is the x component of that distance and ##\hat y## is the y component of it

I just want to start learning some real electricity and magnetism, mainly for fun. I plan to take the electricity and magnetism physics course next semester. So I started with this lecture PDF and I just want to get unstuck. My next physics course won't require calculus, but I want to learn the calculus based part of the physics too so I can eventually understand Maxwell's equations.

## What is a line integral for electromagnetic force?

A line integral for electromagnetic force is a mathematical concept used in the study of electromagnetism. It involves calculating the work done by an electric or magnetic field on a moving charged particle along a given path.

## How is a line integral for electromagnetic force calculated?

To calculate a line integral for electromagnetic force, you must first determine the electric or magnetic field along the given path. Then, you integrate the dot product of the field and the path, taking into account the direction and magnitude of the field at each point along the path.

## What is the significance of line integral for electromagnetic force?

The line integral for electromagnetic force is important in understanding the interaction between charged particles and electric or magnetic fields. It helps in predicting the motion of particles in these fields and is essential in many practical applications, such as in designing electrical circuits and motors.

## What are some real-world examples of line integral for electromagnetic force?

Some examples of real-world applications of line integral for electromagnetic force include calculating the force exerted on a charged particle in a magnetic field, determining the energy required to move a charged particle through an electric field, and predicting the motion of a charged particle in a circuit.

## What other concepts are related to line integral for electromagnetic force?

Other related concepts include electric and magnetic fields, Coulomb's law, Lorentz force, and electromagnetic induction. Understanding these concepts is essential in accurately calculating and interpreting line integrals for electromagnetic force.