Wire going through a magnetic field

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic force acting on a wire carrying a current of 1.5 A as it passes through a 2.0 T magnetic field, making a quarter-circle turn with a radius of 1 meter. The magnetic force is determined using the formula Fb = IL x B, resulting in a magnitude of 3 N. The participants emphasize the importance of expressing the force as a vector and correctly applying the right-hand rule to determine the direction of the force, which can vary based on the orientation of the magnetic field.

PREREQUISITES
  • Understanding of magnetic force on current-carrying conductors
  • Familiarity with the right-hand rule for vector direction
  • Knowledge of the formula Fb = IL x B
  • Basic geometry for calculating lengths of circular arcs
NEXT STEPS
  • Study the application of the right-hand rule in various magnetic field orientations
  • Learn about the effects of magnetic field direction on force vectors
  • Explore the integration of forces on curved wires in magnetic fields
  • Investigate the implications of different current strengths on magnetic force calculations
USEFUL FOR

Physics students, educators, and anyone interested in electromagnetism, particularly those studying the behavior of current-carrying wires in magnetic fields.

timnswede
Messages
100
Reaction score
0

Homework Statement


A wire carrying 1.5 A passes through a region containing a 2.0 T magnetic field. The wire is perpendicular to the field and makes a quarter-circle turn of radius R= 1/
gif.latex?%5Csqrt%7B2%7D.gif
as it passes through the field region as shown below. Find the magnitude and direction of the magnetic force on this section of wire. Use the coordinate system shown and take the +z-axis out of the page. (Hint: Do Not Integrate)

Homework Equations


Fb=ILxB

The Attempt at a Solution


So what I did was I drew a straight line from where the wire enters the field to where it exits. Then I used the Pythagorean theorem to get the length of the wire, which is 1m (assuming meters, could call it units I suppose). Then I used the right hand rule, with ILxB, and got Fb being 90 degrees up and to the right of the wire. So Fb=ILBsin90=3 N.Would this be a correct way of doing it?
3uXMkdy.png
 
Physics news on Phys.org
Why did you do that? The length of the wire is obviously not that of your 'straight wire'. And the force on the wire is a vector so you need to express it that way.
 
Last edited by a moderator:
rude man said:
Why did you do that? The length of the wire is obviously not that of your 'straight wire'. And the force on the wire is a vector so you need to express it that way.
I did it because if I take the integral of a curved wire in a uniform magnetic field it is just the displacement, like in this example http://i.imgur.com/mNSSko3.png. Would that not apply here? I know I can get the length of the wire since it is a quarter circle, but why would the other way not work?
 
timnswede said:
I did it because if I take the integral of a curved wire in a uniform magnetic field it is just the displacement, like in this example http://i.imgur.com/mNSSko3.png. Would that not apply here? I know I can get the length of the wire since it is a quarter circle, but why would the other way not work?
OK, I see what you did. Interesting apprtoach. But you need to express the force as a vector. You have correctly obtained the magnitude but not the complete vector force.
 
rude man said:
OK, I see what you did. Interesting apprtoach. But you need to express the force as a vector. You have correctly obtained the magnitude but not the complete vector force.
Since Fb is perpendicular to my straight line, it would be 45 degrees above the x-axis right?
 
timnswede said:
Since Fb is perpendicular to my straight line, it would be 45 degrees above the x-axis right?
Right.
Would you have gotten the right answer if the B field had pointed in the direction of +z instead of -z? How?
 
rude man said:
Right.
Would you have gotten the right answer if the B field had pointed in the direction of +z instead of -z? How?
Then the Fb is still perpendicular to my straight line, but now it's is below it. So it would be the same magnitude, but 135 degrees below the x axis.
 
timnswede said:
Then the Fb is still perpendicular to my straight line, but now it's is below it. So it would be the same magnitude, but 135 degrees below the x axis.
Right again. I think you have this subject well in hand.
 
  • Like
Likes   Reactions: timnswede

Similar threads

Replies
16
Views
1K
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
14
Views
3K
Replies
4
Views
1K
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
2K