Determining Forces on a Current Loop in a Magnetic Field

[sweetdion]

Homework Statement

A current loop that lines in the horizontal plane consists of a straight line segment 2R and semicircular segment with a radius R. A current I flows through the loop (assume counterclockwise direction), and a uniform magnetic field B points outward of the page everywhere. Provide magnitude and direction for all forces.

*Sorry the picture is so big an blurry*

a) What is the force on straight segment due to magnetic field?b) What is the force on two elements of current dI1 and dI2 symmetrically arranged on the semicircle (sorry I don't know how to draw them, but they are symmertrical to the origin.

Homework Equations

F=ILBsintheta
F=qvBsintheta
*right hand rule*

The Attempt at a Solution

a) since no numbers were given:
L = 2R
F=2IRBsintheta
but I don't know B or theta
and using the right hand rule, the direction of the force is down.

b) This one is different, because there are two elements of current and it is curved. Using the right hand rule, the direction is out of the paper. The equation used to calculate force would be F=qvBsintheta. Which causes the same problem I had in part A.

 

[tiny-tim]

A current loop that lines in the horizontal plane consists of a straight line segment 2R and semicircular segment with a radius R. A current I flows through the loop (assume counterclockwise direction), and a uniform magnetic field B points outward of the page everywhere. Provide magnitude and direction for all forces.
…
a) What is the force on straight segment due to magnetic field?

b) What is the force on two elements of current dI1 and dI2 symmetrically arranged on the semicircle (sorry I don't know how to draw them, but they are symmertrical to the origin.
…
L = 2R
F=2IRBsintheta
but I don't know B or theta
and using the right hand rule, the direction of the force is down.

b) This one is different, because there are two elements of current and it is curved. Using the right hand rule, the direction is out of the paper. The equation used to calculate force would be F=qvBsintheta. Which causes the same problem I had in part A

Hi sweetdion!

(have a theta: θ )

θ is the angle between the directions of the current (I) and of the field lines (B).

This is because the equation F= ILBsinθ is really the vector equation F = LIxB (or is it -LIxB ? ) …

just like the Lorentz force law F = qvxB.

(and you do know B … the question says it's just "B" )

 

[sweetdion]

So for b it doesn't matter that there are 2 current pieces on a curve?

 

[tiny-tim]

So for b it doesn't matter that there are 2 current pieces on a curve?

Hi sweetdion!

I think you're meant to give the force on each piece separately.

If not, find them separately, and add them.

 

[sweetdion]

Viola!

So for part a i get F=2RIB
Part b i get F1=RdlBsinθ and F2=RdlBsinθ

 

[tiny-tim]

Viola!

So for part a i get F=2RIB
Part b i get F1=RdlBsinθ and F2=RdlBsinθ

Who's Viola??

a is the correct magnitude, but you haven't given the direction of the force.

In b, why are you using sinθ?

And isn't dl already a length, so you don't need R?

(and why are you putting dl in italics?)

 

[sweetdion]

Who's Viola??

a is the correct magnitude, but you haven't given the direction of the force.

In b, why are you using sinθ?

And isn't dl already a length, so you don't need R?

(and why are you putting dl in italics?)

voilà means yes! in french. I spelled it wrong the first time.

to give the direction of the force i use the right hand rule. I put my thumb along the current and curl my fingers in the direction of B to give that the force is to the right? Or is there some other way to figure this out.

So for part b I get F=dlB (for one of the pieces of current). And do the same thing for the direction of the force (right hand rule).

sorry, i only put dl in italics because it was italicized in the picture :)

 

[sweetdion]

Sorry i mean its this accordind to wikipedia: The right-hand rule: Pointing the thumb of the right hand in the direction of the conventional current or moving positive charge and the fingers in the direction of the B-field the force on the current points out of the palm. The force is reversed for a negative charge.

So for a i get out of the page and for b i get out of the page again.

 

[tiny-tim]

Hi sweetdion!

(just got up :zzz: …)

to give the direction of the force i use the right hand rule. I put my thumb along the current and curl my fingers in the direction of B to give that the force is to the right? Or is there some other way to figure this out.

So for part b I get F=dlB (for one of the pieces of current). And do the same thing for the direction of the force (right hand rule).

Sorry i mean its this accordind to wikipedia: The right-hand rule: Pointing the thumb of the right hand in the direction of the conventional current or moving positive charge and the fingers in the direction of the B-field the force on the current points out of the palm. The force is reversed for a negative charge.

So for a i get out of the page and for b i get out of the page again.

How can they be out of the page? That's parallel to B, but the force must be perpendicular to B.

What are you doing with your fingers??

Point the thumb and index finger along the two vectors you want to "cross" (they don't have to be at a right-angle) … and point the remaining fingers perpendicular to the palm …

then the cross-product is in the direction of those fingers (ie the same as the normal to the palm).

(btw, I've read the question again, and I'm now sure that when it says "What is the force on two elements of current dI1 and dI2 symmetrically arranged …" it means the total (resultant) force on the pair of elements … I know that looks a bit weird at first sight, but I expect they're thinking of using the result to integrate over all elements )

 

[sweetdion]

Thanks so much for your help, tiny-tim!