Magnetic Fields, Current, and Tension

In summary, the student is stuck on a physics problem and is looking for guidance. He asks for help, but does not want the answer spoon-fed to him. He suggests that the current is going out of the page and that the magnetic field might be the dominating force. He tries to solve for * using the right hand rule and theLaplace force, but is stuck. He asks for help with solving the problem, and notes that he would be more helpful if he could understand why he is getting stuck.
  • #1
twotaileddemon
260
0
Hi everyone. I'm stuck on a physics problem and it's gotten me pretty confused.. and I came here wondering if anyone can guide me without giving me the answer and allowing me to learn the material instead of just receiving the answer without a clue as how to do the problem. I'd greatly appreciate any help ^^;; ... and thanks for reading this !

Homework Statement



A horizontal wire 0.20 m long and 80 grams in mass is hung in a uniform B-field by two massless strings. The magnitude of the magnetic field is 0.070 T. When a 42 amp current passes through the wire, the wire swings upward forming an angle *.

a. What direction is the current going through the wire?
b. What does * equal?
c. What is the tension in each of the two strings?

Diagram: http://img.photobucket.com/albums/v696/talimtails/pp22A.jpg
* = theta

Homework Equations


F = BILsin*
B = kI/r (possibly)
k (constant)= 2 x 10^-7 T-m/A (T = tesla, A = amps, m = meters)
+Application of the right hand rule, with thumb being current, other fingers being magnetic field, and force coming out of the palm.

The Attempt at a Solution


For a, I said the current was going out of the page. I figured the magnetic field would be going down (N to S on the diagram) because magnetic fields usually are directed from + to -. So using my right hand, if the magnetic field (my four fingers excluding thumb) is down, the current (my thumb) seemed to be going out of the page towards me.

For b, I thought there must be some kind of acceleration that is making the wire move. If so, F = ma, but F also equals BILsin*. Therefore, I set the two equal to each other and solved for *.
sin* = ma/BIL = (.08 kg)(9.8 m/s^2)/(.07 T)(42 A)(.2m) ... * = 4/3.
However, I quickly realized that 4/3 is not a possible solution for * as it can never go over 1 for sin. This is where I get stuck. I'm not sure, to be truthful, that acceleration is a factor since it seems the magnetic field would be forcing the wire to the N pole rather than acceleration... but I could not find any other applicable equation. Is there some obvious factor I'm missing? Or.. if you'd rather not give that, how should I go about finding the angle (without giving me a direct answer). I'd be helpful if I could understand why and then do the calculations myself rather than being giving an answer.

For c, recalling early chapters, I would draw a free body diagram of the components I have and solve for T using that. In that case, would I consider the normal force (F_n) and the weight (mg)? I've never done this with magnetic fields before, so it leads me to believe that I should also consider the magnetic field.. as that seems to be the dominating force that determines the angle and therefore tension.
 
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  • #2
Um... ^^; I guess I'll sleep on this.. but any help would still be appreciated :)
 
  • #3
Some notes (trying to solve this myself):

a) correct. Simple application of the Right-Hand rule + Laplace force

b)I understand from the problem that when the wire form an angle θ, it will balance at that position. So draw the forces acting on it:
1)Weight
2)The (2) tensions from the two strings
3)The Laplace force

Note that the laplace force is always horizontal to the "walls" N and S.
Also the following is not correct: F=IBLsinθ. because the angle between the current`s direction and magnetic field`s direction is π/2. So, F=IBL.
Thats why u got sin > 1.

After drawing the forces, take the balance equations.

Hope i helped a little
 
  • #4
I didn't know what the laplace force was.. so I looked it up and got the equation F = q(E + v x B) .. but seeing as how I don't have a velocity, electric field, or charge.. I don't know why I'm supposed to use it.

Weight would be acting down, and tension would be acting on the strings.. but since tension is unknown, I don't know how to find the angle with only one variable.

How do you know the angle between the magnetic field and current's direction is pi/2? And even if you only use IBL, how are you supposed to find an angle without an angle being used in the equation? Unless... you mean to say that I should divide IBL by mg to get (42 A)(.07 T)(.2 m) / (.08 kg)(9.8 m/s^2) ... which would give me .75 and make * (theta) 48.5*...
(I did that, because I drew a free body diagram with the force and weight, and solved for theta... would that work?)
 
  • #5
If anyone wants me to put the diagram I drew on paint I will.. :)
 
  • #6
Well you should study first before trying to solve problems..

Laplace force is (if E=0): F=q*vxB and if you multiply and divide by "t" (time) you get: F=(q/t)*(vt)xB => F=I*LxB or dF=I*dLxB.

Using the right-hand rule you get the force`s direction.
Suppose the magnetic field goes from S to N or N to S. The current`s direction is perpendicular to B. Its easy to see from your diagram..
So: F=I*L*B*sin(π/2)=IBL.
Thats how you get the formula..
 
  • #7
Well I didn't know about Laplace force because we never went over that in class.. and it wasn't discussed in the chapter. So I had no idea to even begin using it. I do take my studies very seriously.. so please don't say or think I don't study.
sin of pi/2 is 1, and then you can get IBL.. I see that.
I was just wondering, since I know force is going to the right due to the right hand rule, would it be correct to divide the F I get from BIL by mg.

See.. I did a FBD (free body diagram).
Force is going to the right because of the right hand rule. We know that mg is going down. so sin* = oppsite / adjacent = (42 A)(.07 T)(.2 m) / (.08 kg)(9.8 m/s^2)
And then I got * = 48.5*
 
  • #8
Sorry, i didn`t want to insult you or anything.. I just noticed that you trying to solve a problem without knowing its theory. Thats all.

I don`t think that it`s correct to say: sinθ=FBL/mg. Show me your FBD if you can please.
 
  • #9
From the balance equations you get:

T*cosθ=mg [1]
T*sinθ=BIL [2]

Thats all.
 
  • #11
The Laplace force (aka Lorentz force) is just the name for the force youre describing with the right hand rule and F=ILBsintheta. In this case, the current and field are at a right angle. JK423's equations are what you need, but remember that the tension is split between 2 strings. Your FBD is okay, but if you want the forces to sum to zero, your tension forces need to be in the opposite direction
 
  • #12
So instead of the way I drew tension in the diagram, it should be in the opposite direction.. right?
And since tension is in two strings, I divide the answer I get for tension by 2.
Now.. if my FBD is okay, then does that mean I got the angle correct?
 
  • #13
I don't think you can do that. The direction of the string is not necesserily(oups :P) the same with the ΣF of mg and BIL.
Even if you could do that(which you cannot) it would be tanθ=f/mg.
Draw all the forces (+ Tension), and get ΣF=0 on the axis which is best. You`ll get the equation [1] and [2] as i wrote above
 
Last edited:
  • #14
T*cosθ=mg [1]
T*sinθ=BIL [2]

I don't understand how you got those equations.. sin is associated with y axis, and cos is associated with the x axis... so wouldn't the cos and sin in your equations be switched?
 
  • #16
Oh.. you don't have to do that oo;! I don't want to trouble you so much.. =(
 
  • #17
twotaileddemon said:
T*cosθ=mg [1]
T*sinθ=BIL [2]

I don't understand how you got those equations.. sin is associated with y axis, and cos is associated with the x axis.

That is the wrong way of thinking. Rules like that only work for angles in standard position. Youve got a triangle, go back to SOH CAH TOA and youll see how he got those equations
 
  • #18
Ohh... I understand now.:)
 
  • #19
So I can solve for the angle using this:
mg/cosθ = BIL/sinθ
mgtanθ = BIL
tanθ = BIL/mg
and.. solve for θ that way..
 
  • #21
I get it now.. thanks :)!
 
  • #22
You really helped a lot.. I understand why now.. =)
 

1. What is a magnetic field?

A magnetic field is an invisible force field that is created by moving electric charges. It is also produced by magnetic materials such as magnets and electric currents.

2. How is a magnetic field created?

A magnetic field is created when electric charges move, either in the form of electric current or by the spinning of electrons in an atom. The movement of these charges creates a magnetic field around the charged object.

3. What is the relationship between magnetic fields and electric currents?

Magnetic fields and electric currents are closely related. An electric current creates a magnetic field, and a changing magnetic field can induce an electric current. This relationship is known as electromagnetism and is a fundamental principle in physics.

4. How does tension affect magnetic fields?

Tension, or the force applied to an object, does not directly affect magnetic fields. However, tension can indirectly impact magnetic fields by changing the position or movement of electric charges, which in turn can alter the magnetic field around the charges.

5. How are magnetic fields and electric fields different?

Magnetic fields and electric fields are different in several ways. Electric fields are created by stationary electric charges, while magnetic fields are created by moving electric charges. Electric fields act on electrically charged objects, while magnetic fields act on magnetic materials or moving electric charges. Electric fields are also stronger than magnetic fields, but they can be converted into each other through the process of electromagnetic induction.

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