Solving for F: A Visual Approach

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Homework Help Overview

The discussion revolves around finding a vector F based on a visual representation provided in an image link. The problem appears to involve concepts related to vector fields and their magnitudes in a two-dimensional plane, particularly in relation to the z-axis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various expressions for the vector F, with attempts to adjust the magnitude based on the distance from the z-axis. Questions arise regarding the correctness of the proposed expressions and their magnitudes.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's attempts. Some guidance has been offered regarding the necessary adjustments to the magnitude of the vector, but no consensus has been reached on a final expression for F.

Contextual Notes

There is a recurring emphasis on the relationship between the vector's components and their magnitudes in relation to the distance from the z-axis, with specific reference to the factor of 4 in the magnitude. Participants are also grappling with the implications of their mathematical manipulations.

-EquinoX-
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Homework Statement



http://img245.imageshack.us/img245/7535/38187200.th.jpg

I need to find F from the picture above

Homework Equations


The Attempt at a Solution



My answer so far is:

\frac{-y}{(x^2+y^2)} \vec{i} + \frac{x}{(x^2+y^2)} \vec{j}

is this correct?
 
Last edited by a moderator:
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No. The magnitude is at a distance r from the z-axis should be 4r. recall that r=\sqrt{x^2+y^2}. Note that your "answer so far" has the properties depicted except that it has magnitude 1/r at a distance r from the z-axis.
 
so it should then be:

\frac{-y}{4(x^2+y^2)} \vec{i} + \frac{x}{4(x^2+y^2)} \vec{j}
 
Does \sqrt{\left(\frac{-y}{4(x^2+y^2)}\right)^2 + \left(\frac{x}{4(x^2+y^2)}\right)^2}=4\sqrt{x^2+y^2}?

No. Look, -y\vec{i}+x\vec{j} is a vector field whose vectors point in all the right directions, make them the proper length.
 
I have no clue... I've tried -4y\vec{i} + 4x\vec{j} and it seems to be wrong as well
 
-EquinoX- said:

Homework Statement



http://img245.imageshack.us/img245/7535/38187200.th.jpg

I need to find F from the picture above

Homework Equations





The Attempt at a Solution



My answer so far is:

\frac{-y}{(x^2+y^2)} \vec{i} + \frac{x}{(x^2+y^2)} \vec{j}

is this correct?

benorin said:
No. The magnitude is at a distance r from the z-axis should be 4r. recall that r=\sqrt{x^2+y^2}. Note that your "answer so far" has the properties depicted except that it has magnitude 1/r at a distance r from the z-axis.

-EquinoX- said:
so it should then be:

\frac{-y}{4(x^2+y^2)} \vec{i} + \frac{x}{4(x^2+y^2)} \vec{j}
As benorin said, your first attempt had magnitude 1/r and it should be 4r. For your second attempt you just divided by 4. That makes the magnitude 1/4r, not 4r. You need to multiply your first answer by 4r2= 4(x2+ y2).
 
Last edited by a moderator:
so it's just simply:

-4y\vec{i} + 4x\vec{j}
 

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