Magnetic force for a current-carrying wire

AI Thread Summary
The discussion revolves around calculating the magnetic force on a current-carrying wire in the presence of the Earth's magnetic field. The problem involves a 250-meter wire carrying a current of 110 A, with the magnetic field directed north at 72 degrees below the horizontal. Participants express confusion about the correct setup of the magnetic field and the angles involved in the calculations. The force direction is clarified, indicating that for part A, the force acts into the page, while for part B, the angle remains 72 degrees as the current direction changes to south. The conversation emphasizes the importance of understanding the geometry of the magnetic field and current orientation for accurate force calculations.
CaneAA
Messages
13
Reaction score
0

Homework Statement



A high-voltage power lines carries a current of 110 A at a location where the Earth's magnetic field has a magnitude of .59 G and points to the north, 72 degrees below the horizontal. Find the direction and magnitude of the magnetic force exerted on a 250-m length of wire if the current in the wire flows (a) horizontally toward the east or (b) horizontally toward the south.

Homework Equations



F=ILBsintheta

The Attempt at a Solution



What confuses me about this problem is setting up the magnetic field "pointing north, 72 degrees below the horizontal." I've attached a sketch, I don't know if it is correct.

If it is then the F=(110)(250)(5.9*10^-5)sin72 = 1.54 N

And with the 2nd RHR I get that the force is pointing in the positive z direction.

For part B, I don't know how the sketch would change. The current is still "horizontal", but this time, the horizontal direction points the the south. Does the angle with the current stay the same?
 

Attachments

  • 1.jpg
    1.jpg
    4.3 KB · Views: 1,434
Physics news on Phys.org
Hi CaneAA! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)
CaneAA said:
What confuses me about this problem is setting up the magnetic field "pointing north, 72 degrees below the horizontal." I've attached a sketch, I don't know if it is correct.

Yes, except that your arrow is the wrong way round …

"72 degrees below the horizontal" means that the field (and the arrow) points downward (it's not like wind direction! :wink:)
For part B, I don't know how the sketch would change. The current is still "horizontal", but this time, the horizontal direction points the the south. Does the angle with the current stay the same?

Nooo!

A sketch doesn't help much in these 3D situations: it doesn't tell you what θ is! :redface:

I prefer to use the unit vectors i j and k,

with the formulas i x j = k, j x k = i, k x i = j :smile:
 
Hi Tiny-tim, thanks for your reply.

So for part A, the angle would be 72, but the force would be directed straight into the page (-z direction). Is this correct?

For part B, I'm having a hard time following your answer since my physics class doesn't go into cross product--we find direction and such using right hand rule. Is there another way to do this problem than the one you mentioned? (Why wouldn't do the angle also be 72?)

Thank you :smile:
 

Attachments

  • part1.jpg
    part1.jpg
    4.7 KB · Views: 1,314
  • partb.jpg
    partb.jpg
    4.8 KB · Views: 1,512
Hi CaneAA! :smile:

Actually, now I take a second look, the angles are particularly easy in this case.

Let's have a look:
CaneAA said:
So for part A, the angle would be 72, but the force would be directed straight into the page (-z direction). Is this correct?

No, For part A, B is in the vertical north (x,z) plane, and L is east, which is perpendicular to that plane, isn't it? :wink:

For part B, L is south, which is in the same plane, so yes the angle is 72°.
 
I don't understand what you're trying to tell me. :confused:

For part A, why do you say it's perpendicular? I see that the B and I are on separate planes, but isn't the angle between the current and the magnetic field 72? *look at my (beautiful) hand-drawing.
The only other option I see for \vartheta (<-- look I learned.) is the supplement of 72. But don't you always choose the smaller angle between B and I?
 

Attachments

  • 12.jpg
    12.jpg
    15.3 KB · Views: 1,346
CaneAA said:
… the Earth's magnetic field has a magnitude of .59 G and points to the north, 72 degrees below the horizontal. Find the direction and magnitude of the magnetic force exerted on a 250-m length of wire if the current in the wire flows (a) horizontally toward the east
tiny-tim said:
No, For part A, B is in the vertical north (x,z) plane, and L is east, which is perpendicular to that plane, isn't it?
CaneAA said:
I don't understand what you're trying to tell me. :confused:

For part A, why do you say it's perpendicular? I see that the B and I are on separate planes, but isn't the angle between the current and the magnetic field 72? *look at my (beautiful) hand-drawing.
The only other option I see for \vartheta (<-- look I learned.) is the supplement of 72. But don't you always choose the smaller angle between B and I?

(btw, I meant the θ you can copy and paste from my signature, without using latex :wink:)

Stand facing North, and stick out your right hand … that's East …

the magnetic field is in the vertical plane in front of you, and your right arm is the normal to that plane …

so it's perpendicular to every line in that plane (including the magnetic field). :smile:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top