Work done by electromagnetic force

AI Thread Summary
To calculate the work done by the force F along the path y=x² from (0,0) to (2,4), a line integral must be performed, requiring the dot product of the force vector F and the displacement vector dl. The force is not conservative if the work done depends on the path taken, which is clarified by the distinction between open and closed paths. A conservative force's work is independent of the path, relying only on the endpoints, and can be expressed as the gradient of a potential function. The discussion emphasizes the importance of correctly defining the infinitesimal displacement vector dl in Cartesian coordinates. Understanding these principles is crucial for solving the problem accurately.
scarlets99
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Homework Statement



1.Calculate the work done by the force F=i cos(pix/4) + j y2 along the path y=x2 in the x-y plane from (0,0) to (2,4).

2.Is the force F conservative? Explain.

Homework Equations



Work = Integration between a and b of F.dl

The Attempt at a Solution



1. I know it's a line integral between the 2 points, I'm just unsure of how to do it.

2.If W=0 then it is conservative
 
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scarlets99 said:
1. I know it's a line integral between the 2 points, I'm just unsure of how to do it.

You have to find the dot product of the vector F and the vector dl and then integrate the result over the curve.

scarlets99 said:
2.If W=0 then it is conservative

Nope. Think about this, does it make sense that a conservative force never does work between any two points? (Gravity and the electrostatic force are both conservative).

You're probably thinking of a result that the line integral of a conservative vector field around a closed path is 0. But in this case it's not a closed path, since the starting and ending points are not the same.

It's a conservative force if the work done in moving between any two points is independent of the path taken between them. In other words, all line integrals between point A and B, regardless of path, give you the same result. So the work done depends only on the two endpoints (you can probably see how the thing about closed paths follows from this property).

Another way to think about it is that a conservative vector field can always be expressed as the gradient of some scalar function usually called a "potential function" (which in the case of conservative forces actually represents the potential energy). So you can figure out the work done between any two points just by looking at the difference in potential between them. It doesn't matter how you got from one to the other.
 
Thank you cepheid.
What is the value of dl?
 
scarlets99 said:
Thank you cepheid.
What is the value of dl?

Well, I think that moving a small (infinitesimal) displacement dl along the curve means moving a small displacement in the x-direction + a small displacement in the y-direction. Does that help? (How would you express what I just said in vector form?)
 
dxdy?
So the integral between 0 and 2, and 0 and 4 F.dxdy
 
scarlets99 said:
dxdy?

NO. dxdy has dimensions of area (m2), so you know that it can't possibly be a displacement vector.

I'm not sure why you multiplied. Take a look again at what I said :

cepheid said:
Well, I think that moving a small (infinitesimal) displacement dl along the curve means moving a small displacement in the x-direction + a small displacement in the y-direction. Does that help? (How would you express what I just said in vector form?)

Another hint: it might help to think about a simpler example first: if I move from the origin 3 m in the x-direction and then 2 m in the y-direction, how would I express that displacement in Cartesian vector form? Now, instead of 3 m and 2 m, generalize that to infinitesimal straight-line displacements dx and dy (infinitesimal since you're actually moving along a curve and just approximating it as a sequence of straight-line moves, and then taking the limit as the size of those straight-line moves goes to zero).
 
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