# Simple loop question

1. May 22, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
Problem 5.4 Suppose that the magnetic field in some region has the form $B=kz \hat{x}$ where k is a constant). Find the force on a square loop (side a), lying in the yz plane and centered at the origin, if it carries a current I, flowing counterclockwise, when you look down the x axis.

2. Relevant equations
$$\vec{F} = I \int d\vec{l} \times \vec{B}$$

3. The attempt at a solution
Now i can easily prove that the forces in the two sides that are in line with tehe Y axis cancel out

For the segments parallel to the Z axis however things are a bit different
on the top segment for z>0
$$F = i \frac{a}{2} k \frac{a}{2} (\hat{-z}\times\hat{x}) = \frac{ika^2}{4} \hat{y}$$
For the bottom segment z>0
$$F = i \frac{a}{2} k \frac{a}{2} (\hat{z}\times\hat{x}) = -\frac{ika^2}{4} \hat{y}$$

For the top segment z<0
$$F = i \frac{a}{2} k \frac{-a}{2} (\hat{-z}\times\hat{x}) = -\frac{ika^2}{4} \hat{y}$$
For the bottom segment z<0
$$F = i \frac{a}{2} k \frac{-a}{2} (\hat{z}\times\hat{x}) = \frac{ika^2}{4} \hat{y}$$

but dont all these forces jsut cancel to zero??

where have i gone wrong?? I know the answer cant be zero because this is not a uniform field...

(edit: i have changed the diagram since i had done it wrong, the current should have been in the opposite direction)

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Last edited: May 23, 2007
2. May 23, 2007

### Meir Achuz

The forces on the top wire and the bottom wire are in the same direction.
z is only >0 for the top wire and only <0 for the bottom wire.
z can't be <0 at the topp wire.
There are only two forces to add.

3. May 23, 2007

### stunner5000pt

why cant z <0 for hte top wire... doesnt part of the wire lie in teh z<0 region? And isnt the field different in taht region aswell??

4. May 23, 2007

### Staff: Mentor

Yes.

5. May 23, 2007

### stunner5000pt

on second thought... it doenst cancel out

for hte right side loop
$$F = i \frac{a}{2} \hat{y} \times k \frac{a}{2} \hat{x} = \frac{ika^2}{4} \hat{z}$$

$$F = i \frac{a}{2} (-\hat{y}) \times k \frac{-a}{2} \hat {x} = \frac{ika^2}{4} \hat{z}$$

so the force points to the right, according to my diagram and has value $$\vec{F} = \frac{ika^2}{2} \hat{z}$$

thanks a lot!!

Last edited: May 23, 2007
6. Mar 27, 2011

### Brown Arrow

when i do the Right hand rule i see that on top force is up and at bottom its down so they cancel
now the sides the force also points in opposite direction so they cancel....
this does not make sense to me at all could some1 explain plz

Last edited: Mar 27, 2011
7. Mar 27, 2011

### ideasrule

Are you sure you got the directions of the magnetic field right? To summarize, if k>0, the field points up on the right side of the image and down on the left side.