What Are the Correct Directions for the Forces in These Diagrams?

AI Thread Summary
The discussion focuses on determining the correct directions of forces on wires in four provided diagrams using the right hand rule and the formula F = Il x B. For diagram 1, the force is identified as UP; for diagram 2, it is OUT; for diagram 3, the force is DOWN; and for diagram 4, the force is IN. Participants emphasize the importance of correctly applying the right hand rule to avoid errors in direction. The conversation concludes with a reminder that using the right hand is crucial for accurate results.
shimizua
Messages
99
Reaction score
0

Homework Statement


Decide from the following list of possibilities what is the appropriate direction of the force on the wire for each of the diagrams 1 through 4:


A: up; B: down; C: left; D: right; E: into the page; F: out of the page; G: None of these; H: No force


Input the answers for diagrams 1-4 as they are posed - no spaces, no commas. For example if the force on diagram 3 is into the page and there is no force on any of the other wires then input 'HHEH'.
1.
http://capa-new.colorado.edu/teacher/capalibrary/Graphics/Gtype61/prob20v8.gif
2.
http://capa-new.colorado.edu/teacher/capalibrary/Graphics/Gtype61/prob20v1.gif
3.
http://capa-new.colorado.edu/teacher/capalibrary/Graphics/Gtype61/prob20v3.gif
4.
http://capa-new.colorado.edu/teacher/capalibrary/Graphics/Gtype61/prob20v6.gif

Homework Equations


well i used the right hand rule and came up with BAEG but don't know what i did wrong. Not too good with the right hand rule


The Attempt at a Solution

 
Physics news on Phys.org
This is the way I learned it: F = IlxB , where I is the current, l is the length (direction too, since it is a vector quantity), and B is the magnetic field, with the cross product operation.

So for figure 1: point the fingers of your right hand in the direction of I, and point your palm in the direction of B. The direction of your thumb will be the direction of the force. So the force will be UP.

Figure 2: fingers in the direction of I, palm in the direction of B, so the force is OUT.

Figure 3: fingers in the direction of I, palm in the direction of B, force points in the direction of the thumb: the force is DOWN.

Figure 4: fingers of the right hand in the direction of I, palm in the direction of B, thumb points in the direction of the force: so the force is IN.

Alternatively, you can think of it as the direction a screw would advance if you put the tails of the two vectors together, and screwed I in the direction of B through the smaller angle.

Hope this helps.
 
hey thanks that was right. i really hate right hand rule stuff
 
It's really simple stuff. Just point your fingers and palm in the right direction, and you've got the direction of the force. It all comes from the formula with the cross product.

And always remember to use your right hand, or you'll have the force in the wrong direction. It is the right hand rule, after all.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top