Applying cross product to some problem

[Genericcoder]

Hi,

So I still not sure how to apply like rhr rule in this setup in problem like the one in the following so I tried to do rhr in order to get the direction but it didn't work out. this is an example from halliday and resnick book.


Figure 32-24 shows a wire segment,placed in a uniform magnetic field B that points out of the plane of the figure. If the segment carries a current i, what resultant magnetic force acts on it ?
here is image of the problem at hand.



Solution:

f1 = f3 = -iLB

so its equal to

iL = <iL,0,0>
B = <0,0,B>

iLx B = <0,-iLB,0>
so it points in the negative y direction I tried to do same thing now for circular part to get the directions.

iL = i<rcostheta,rsintheta,0>
B = <0,0,B>

iL x B = <rsinthetaB,rcostheta,0>

so since its a circle costheta will gets canceled by symmetry
so it will be so it will have only a horizontal component equal to rsinthetaB ofcourse this works out in the problem when you do the integeral but its not right reasoning as the book have and book used rhr.

 

[ehild]

Hi,

So I still not sure how to apply like rhr rule in this setup in problem like the one in the following so I tried to do rhr in order to get the direction but it didn't work out. this is an example from halliday and resnick book.


Figure 32-24 shows a wire segment,placed in a uniform magnetic field B that points out of the plane of the figure. If the segment carries a current i, what resultant magnetic force acts on it ?
here is image of the problem at hand.



Solution:

f1 = f3 = -iLB

so its equal to

iL = <iL,0,0>
B = <0,0,B>

iLx B = <0,-iLB,0>
so it points in the negative y direction

What are f1 and f3? Do you mean L is the length of the straight wire segments?

I tried to do same thing now for circular part to get the directions.
B = <0,0,B>

iL x B = <rsinthetaB,rcostheta,0>

iL = i<rcostheta,rsintheta,0>

You have to use an elementary segment of the circle IΔL, to get the force, acting on that element and then integrating for the half circle. IΔL is tangential to the curve. What do you call theta? What are the limits of integration? What are the components of force?

so it will be so it will have only a horizontal component equal to rsinthetaB ofcourse this works out in the problem when you do the integeral but its not right reasoning as the book have and book used rhr.

Along the curved path current I flows from left to right between the ends of the straight wire segments. Imagine a box around the circular part. You see the current flowing in and out, along horizontal direction. ΔL is horizontal, and the effective length is equal to 2R, and you can apply the rhr.

 

[Genericcoder]

yeh I see in that way I u can notice its pointing towards the center but I always want to apply cross product to verify my work in order to see if my rhr rule is correct or not because sometimes I get it wrong coz of human error,but the math of I never get it wrong once I expand using a determinant.
F1,F3 is force along the straight segments I am using the theta to be the angle between the horizontal and the vertical like the following picture.

 

[ehild]

Cross product is all right, but you calculated the magnetic force incorrectly. You need to integrate the forces acting on the infinitesimal pieces of the curved wire, ΔL. What are the components of ΔL at a given theta?

ehild

 

[Genericcoder]

Ye understand I am just trying to get certain element atleast of the integeral like what they did in the book using rhr and using symmetry argument that x component will cancel.iL = <ircos(x),irsin(x),0>
B = <0,0,B>so now applying cross product I will get iLxB = <irsinxB,-ircosxB,0>

but it should be iLxB = <-ircosxB,irsinxB,0>
here we can see the x component of this vector will cancel but I don't know where I am going wrong to get them reversed as I said theta is measured with respect to x-axis.

 

[ehild]

Ye understand I am just trying to get certain element atleast of the integeral like what they did in the book using rhr and using symmetry argument that x component will cancel.iL = <ircos(x),irsin(x),0>

That is wrong. ΔL is tangent to the circle. You wrote the components of the radius-vector, instead of those of IΔLehild

 

[Genericcoder]

Oh I see that's y the angle will from above so we will have
idL = ird(theta) = <irsin(x),ircos(x),0>so crossed we will have
idL = <irsinx,ircos(x),0>
B = <0,0,B>

idL x B = ircos(x) * B j^ + -irsin(x)*Bj^2 is that right?

 

[Genericcoder]

and the integeral will be evulated from PI to 0 so we should have the same answer as book but I was wondering is my reasoning right?

 

[ehild]

write theta instead of x... it is OK then. ehild

 

[Genericcoder]

Thanks a lot ehild yeh I guessed it would be better to write x instead of theta since its smaller sorry I didn't denote that but yeh thanks a lot everything makes sense now.

 

[ehild]

You are welcome

ehild