B Cylinders

1. Mar 6, 2013

Gee Wiz

1. The problem statement, all variables and given/known data
Two very long coaxial cylindrical conductors are shown in cross-section above. The inner cylinder has radius a = 2 cm and caries a total current of I1 = 1.2 A in the positive z-direction (pointing out of the screen). The outer cylinder has an inner radius b = 4 cm, outer radius c = 6 cm and carries a current of I2 = 2.4 A in the negative z-direction (pointing into the screen). You may assume that the current is uniformly distributed over the cross-sectional area of the conductors. What is Bx, the x-component of the magnetic field at point P which is located at a distance r = 5 cm from the origin and makes an angle of 30o with the x-axis? Bx =?

https://www.smartphysics.com/Content/Media/Images/EM/IE/Bcylinders/pic.gif [Broken]

2. Relevant equations

B=(Iμ)/(2πr)

3. The attempt at a solution

Okay so I was able to come up with the correct answer, but I am not sure exactly how. While its nice to get the question right with the snazzy green check mark..it doesn't help in the future. So, I'll try to explain the part where its not clicking for me. I first found the total B at point P. So, i figured i could just use trig to get the x component. I used cosine, which did not yield the correct result, but sine did. I do not understand why i would use sine here. If i treat B as the vector going from the origin to P it looks to me that the correct thing to do would be to use cosine..=/ but its not

Last edited by a moderator: May 6, 2017
2. Mar 6, 2013

TSny

Can you describe the pattern of magnetic field around a long straight current?

3. Mar 6, 2013

Gee Wiz

The b field follows the right hand rule. Point the thumb in the direction of the current and your fingers curl in the direction of B. I know that b isn't going from the origin to the point, or well maybe i dont, but that is how i was trying to picture it

4. Mar 6, 2013

TSny

So, how is the direction of B at point p related to the direction of the radius r?

5. Mar 6, 2013

Gee Wiz

its going either clock or counterclockwise?

6. Mar 6, 2013

TSny

Yes. Using the right-hand rule you should be able to decide which. So, if you were to draw the magnetic field vector (arrow) at point P, what angle would the vector make to the radius r?

7. Mar 6, 2013

Gee Wiz

I believe it would tangent to the radius.

8. Mar 6, 2013

TSny

I'm not sure what "tangent to the radius means". Did you mean tangent to the circle that passes through P and has center at the origin? See http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/magnetic/magcur.html [Broken]

Last edited by a moderator: May 6, 2017
9. Mar 6, 2013

Gee Wiz

I think what i tried to say was said very poorly. Tangent wasn't the correct term to use. I should have said that the b field goes around in a circle (different values of r will yield different results for the strength of the b field) about the current wire.

Last edited by a moderator: May 6, 2017
10. Mar 6, 2013

Gee Wiz

"So, if you were to draw the magnetic field vector (arrow) at point P, what angle would the vector make to the radius r?"

I don't think I answered this. And now i'm not quite sure what i mean. If i draw a vector i think it would be pointing upward and to the left.

Last edited: Mar 6, 2013
11. Mar 6, 2013

TSny

OK. So, the magnetic field line that passes through point P is a circle of radius r and the field line either goes clockwise or counterclockwise around this circle. To decide which, you would need to determine the net amount of current enclosed within that circle (which I assume you already did since you got the correct answer for the magnitude of B). If the net current is out of the page, then the right hand rule will tell you the field line goes counterclockwise. Let's assume that's the case.

Remember that the magnetic field vector at a point is tangent to the magnetic field line through that point. Thus, B is tangent to the circular field line. So, looking at the attached figure, what is the angle between B and r? You need to find the x-component of B, so think about the angle that B makes to the horizontal or vertical.

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12. Mar 6, 2013

Gee Wiz

So B and r make 90 degree angle. And r with the y-axis makes the complement angle of the given angle. If i then make a right triangle using the y-axis as a leg and r as a hypotenuse. That will give a line (the other leg) parallel to the x-axis. So, then the complement of the complement, or just the given is the the angle between this new parallel x-axis line and r. Since r and b make 90 degrees i take the complement again, and this is the angle b makes with the horizontal. So, to conclude this ramble B makes the complement angle with the horizontal. So if i take the cosine of that angle i get the correct answer, or i can take the sin of the given angle. ?

13. Mar 6, 2013

Gee Wiz

https://www.smartphysics.com/Content/Media/Images/EM/IE/Bcylinders/hpic6.gif [Broken]

Last edited by a moderator: May 6, 2017
14. Mar 6, 2013

TSny

Yes. That's it. Think about the sign of the x-component of B.

15. Mar 6, 2013

Gee Wiz

The y component would be positive and x component would be negative

16. Mar 6, 2013

TSny

Good.

17. Mar 6, 2013

Gee Wiz

Sweet! Thank you again for all your help. It really does help me understand it better and more thoroughly so that I am able to better grasp concepts. My specialty is plugging random things into the calculator and getting the right result, but then unable to replicate it. Not a great thing I know, so i am trying to work on that. You've been really helpful on multiple occasions i can't thank you enough.