Solving for F: A Visual Approach

[-EquinoX-]

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]

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-]

so it should then be:

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

 

[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.

 

[-EquinoX-]

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

 

[HallsofIvy]

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?

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).

 

[-EquinoX-]

so it's just simply:

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