Solving Magnetic Torque Problems with Unit Vectors: Homework Help

[kosmocomet]

Homework Statement

Homework Equations

• m=N*I*A
• T=m X B
• m = n(unit vector) *m

The Attempt at a Solution

To calculate m, I know it is just plugging in the information. Thus, m=0.8. Now, the question is computing the unit vector, which has me confused. Using the right hand rule, and going along the current, The normal should look something like this, correct:

If so, how do I make this into an x(unit) and y(unit)?

Any help is much appreciated!

 

[kuruman]

You have ##\vec {\tau}=\vec m \times \vec B##. What is the magnitude of the cross product in terms of the magnitudes ##m##, ##B## and the angle between the magnetic moment and the magnetic field? This is quicker than figuring out unit vectors.

 

[kosmocomet]

You have ##\vec {\tau}=\vec m \times \vec B##. What is the magnitude of the cross product in terms of the magnitudes ##m##, ##B## and the angle between the magnetic moment and the magnetic field? This is quicker than figuring out unit vectors.

Thanks, for the comment. I still am a little confused. The angle between the magnetic field and moment would be 60 degrees correct? If so, how do I apply this for a torque vector

 

[kuruman]

Thanks, for the comment. I still am a little confused. The angle between the magnetic field and moment would be 60 degrees correct? If so, how do I apply this for a torque vector

Yes the angle between ##\vec m## and ##\vec B## is 60^o. There are two ways of finding the cross product as shown here
http://hyperphysics.phy-astr.gsu.edu/hbase/vvec.html

 

[Babadag]

You need to find the n[unit vector] components nx,ny,nz and to multiply with Bx,By,Bz vector
product. The n vector is located in the center of surface A and is equal with the difference between n1 and n2 [end vectors].In order to find the unit vector you have to divide each coordinate by the module=sqrt(x^2+y^2+z^2).[vector product=cross product]

 

[kosmocomet]

Yes the angle between ##\vec m## and ##\vec B## is 60^o. There are two ways of finding the cross product as shown here
http://hyperphysics.phy-astr.gsu.edu/hbase/vvec.html

So I understand how to do the cross product. The problem is I don't know how to define ##\vec m## as an x and y components like B is = Boy(unit vector).

 

[kuruman]

So I understand how to do the cross product. The problem is I don't know how to define ##\vec m## as an x and y components like B is = Boy(unit vector).

Oh, that. This is what you do. (a) Make a better drawing than in Post #1 showing both x and y axes. (b) Draw in ##\hat n##. (c) Resolve ##\hat n## into its x and y components just as you would resolve any vector except that for magnitude you use 1.

 

[kosmocomet]

Oh, that. This is what you do. (a) Make a better drawing than in Post #1 showing both x and y axes. (b) Draw in ##\hat n##. (c) Resolve ##\hat n## into its x and y components just as you would resolve any vector except that for magnitude you use 1.

Hmmm, i think i get it. Thanks.

 

[Babadag]

 

[kuruman]

View attachment 224865

According to your drawing, ##\hat n =\vec {n}_{1y}-\vec {n}_{2y}##. This difference is independent of the choice of origin, so if the task is to find the difference, you might just as well put the origin at point a to make the algebra less involved.

 

[Babadag]

Right. Thank you, kuruman!