# Finding F

## Homework Statement

http://img245.imageshack.us/img245/7535/38187200.th.jpg [Broken]

I need to find F from the picture above

## The Attempt at a Solution

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

is this correct?

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benorin
Homework Helper
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:

$$\frac{-y}{4(x^2+y^2)} \vec{i} + \frac{x}{4(x^2+y^2)} \vec{j}$$

benorin
Homework Helper
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

HallsofIvy
Homework Helper

## Homework Statement

http://img245.imageshack.us/img245/7535/38187200.th.jpg [Broken]

I need to find F from the picture above

## The Attempt at a Solution

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

is this correct?
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:

$$\frac{-y}{4(x^2+y^2)} \vec{i} + \frac{x}{4(x^2+y^2)} \vec{j}$$
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).

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so it's just simply:

$$-4y\vec{i} + 4x\vec{j}$$