Magnetic force for a current-carrying wire

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SUMMARY

The discussion focuses on calculating the magnetic force exerted on a 250-meter length of wire carrying a current of 110 A in the presence of the Earth's magnetic field, which has a magnitude of 0.59 G and is directed 72 degrees below the horizontal toward the north. For part A, the magnetic force is calculated using the formula F = ILBsinθ, resulting in a force of 1.54 N directed into the page (-z direction). For part B, the angle remains 72 degrees, and the current direction changes to south, affecting the force direction but not the angle with the magnetic field.

PREREQUISITES
  • Understanding of magnetic force calculations using F = ILBsinθ
  • Familiarity with the right-hand rule for determining force direction
  • Knowledge of vector components in three-dimensional space
  • Basic concepts of Earth's magnetic field orientation
NEXT STEPS
  • Study the application of the right-hand rule in different current and magnetic field orientations
  • Learn about vector cross products and their relevance in physics
  • Explore the implications of magnetic fields on current-carrying conductors
  • Investigate the effects of varying magnetic field strengths on force calculations
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CaneAA
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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?
 

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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:
 

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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?
 

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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:
 

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