# Finding F

1. Apr 29, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data

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

I need to find F from the picture above

2. Relevant equations

3. 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: May 4, 2017
2. Apr 30, 2009

### benorin

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.

3. Apr 30, 2009

### -EquinoX-

so it should then be:

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

4. Apr 30, 2009

### benorin

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.

5. Apr 30, 2009

### -EquinoX-

I have no clue.... I've tried $$-4y\vec{i} + 4x\vec{j}$$ and it seems to be wrong as well

6. Apr 30, 2009

### HallsofIvy

Staff Emeritus
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: May 4, 2017
7. Apr 30, 2009

### -EquinoX-

so it's just simply:

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