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Homework Help: Torque experienced by a current loop in uniform magnetic field

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Revered members,
    please see my attachment (1)
    In this diagram, PQRS is a rectangular current loop and I is the current and its direction is shown in red arrow . B is the magnetic field and it acts from left to right and n is the normal drawn to the plane of the loop.
    [itex]\Theta[/itex] is the angle between normal to the plane of the loop and the direction of magnetic field.

    2. Relevant equations

    I have to find the angle between I(QR) i.e current element and B, similarly angle between I(SP) and B
    In the book angle between the I(QR) and B is given as (90 - [itex]\Theta[/itex]).
    similarly angle between I(SP) and B is given as ( 90 + [itex]\Theta[/itex])

    3. The attempt at a solution

    since n is normal to the plane of loop, angle between n and SP (and ) n and QR is 90 degree.
    so if angle between n and B is [itex]\Theta[/itex], angle between B and SP (and) B and QR should be 90 - [itex]\Theta[/itex] so that when i add
    90 - [itex]\Theta[/itex] + [itex]\Theta[/itex] i will get 90 degree which is the angle between normal to the loop and concerned current element.
    But my teacher says what is given in the book is correct. Where do i go wrong? Please help revered members. Thanks in advance
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Aug 17, 2011 #2
    This is where you went wrong. Read what you just wrote again. If you were correct then that would be your final answer for both current elements, there is no need to add [itex]\Theta[/itex] again, because you are looking for the angles that each of the two current elements makes with B, which is what you just claimed you found to be [itex]90-\Theta[/itex] for both.

    That is wrong, however you were close. The [itex]90-\Theta[/itex] is actually the answer to one of them, just like the book tells you.
    B lies in between QR and n, where the angle between QR and n is 90, so if you just draw that 2-dimensionally on paper it is obvious that the first answer is just [itex]90-\Theta[/itex] as the book says.

    Now, the direction is important. For SP, notice the direction is different than PS, also the current is in the direction of SP. So you are finding the angle from the other side. So you have [itex]90+\Theta[/itex] instead of minus.

    See attachment if you still don't understand. The red lines are the angles you are looking for.

    Attached Files:

  4. Aug 17, 2011 #3
    i got it sir. Really u have taken painstaking effort to clear my doubt. You have explained what my teacher failed to do. Thanks a lot sir. Thanks again and again
  5. Aug 18, 2011 #4
    Sir, according to your attachments, i have to measure the angle from the point where the current terminates, i.e, when current flows along QR, since it ends at R i have to measure the angle from R and when current flows along SP, since it ends at P, i have to measure from P.Am i interpreted your reply correctly,sir?
  6. Aug 18, 2011 #5
    Right, that's correct.
    Sorry it took me so long to see this.
  7. Aug 18, 2011 #6
    Thank u sir. Is this a rule for measuring angles in current loop?
  8. Aug 19, 2011 #7
    Pretty much.

    If you think about it, the only relevant thing here is the moving charges. This is because of the lorentz force law:
    [tex]\vec{F} = q\left( \vec{E}+\vec{v}\times\vec{B}\right)[/tex]

    So what matters here is going to be the [itex]\vec{v}\times\vec{B}[/itex] and [itex]\vec{v}[/itex] is determined by the direction of the current (i.e. the direction in which the charged particles are moving in is the same as the direction of current). Of course, the actual moving charges are the electrons which go in the opposite direction of the conventional current, but taking this into account is the same as just making q positive.

    Also, because [itex]\vec{B}[/itex] is a vector, you want to find the angle between two vectors, not a vector and a line.
    So what do you use as the direction for the line? The direction of the current is the only thing that makes sense to use as your second vector.
    As I was saying above it's not only because current is really the only thing you have indicating a direction in the first place, but also because the magnetic field will be related to the direction of the current so the direction of the current is the relevant direction that we consider in electromagnetism.
    Last edited: Aug 19, 2011
  9. Aug 19, 2011 #8
    "The direction of the current is the only thing that makes sense to use as your second vector."
    But sir, current is a scalar?
  10. Aug 20, 2011 #9
    Should that matter? Nothing I said really requires current to be a vector anyway.
    If it makes you feel better you can think of the quantity [itex]\vec{J}[/itex] which is the current density, and that is a vector.
    Anyway, my whole argument was based on the direction the "positive" charges of the conventional current were moving, which is [itex]\vec{v}[/itex].

    You are technically right, but you are still going to run into people who say "direction of current" all the time, it's just easier to talk that way but you know what we mean.

    Also I think I mentioned this in my very first post here. If you don't like the argument I'm giving for the direction, then simply read your problem more carefully as it asks for the angle from SP, and not PS.
  11. Aug 21, 2011 #10
    Thanks a lot sir. I understood. You said that we cannot find angle between a line and vector, the direction of current is taken as the second vector. Now the current flows along QR and Magnetic field's direction is along QR, so whats wrong in measuring the angle from the starting point of origination of current, that is from Q, instead of measuring from the terminating point R. Sorry for stretching you, but since you have so many explanations beautifully to this concept, i am asking you to clear all my doubts.
    Last edited: Aug 21, 2011
  12. Aug 21, 2011 #11
    Here's the thing. This is somewhat of a bad question because its wording includes electromagnetism concepts, but really it is just a geometry question.

    So to be honest, if I just looked at this as a geometry problem I would have given the same answer you had in mind originally, that it is [itex]90-\Theta[/itex], because typically when measuring the angle between a line and something else the answer you report is the smallest of the two possible angles because a line doesn't have a direction.

    However, since we know that the answer is [itex]90+\Theta[/itex] I was just attempting to justify this by using concepts from electromagnetism.

    So the point is this, the way the question itself is worded... if I were the teacher I would accept [itex]90-\Theta[/itex] as an answer for both, because it just says what is the angle between some vector and a line segment.

    But if I were the teacher I would also go on to explain to you why I think the book chose that specific angle ([itex]90+\Theta[/itex]) for SP. Even though either angle is technically correct, the book is looking ahead, it's looking at what angle you would be finding if this was a real electromagnetism problem. Considering that you would be looking at q, [itex]\vec{v}[/itex], and [itex]\vec{B}[/itex]. But, keeping in mind that we are dealing with the conventional current so we would have q>0 and therefore in the line segment SP, [itex]\vec{v}[/itex] would be in the direction of S to P.
    So, now the angle between [itex]\vec{v}[/itex] and [itex]\vec{B}[/itex] is definitely [itex]90+\Theta[/itex].

    However, keep in mind that even this is completely arbitrary, you could just as easily have used the real charge carriers (the electrons) which means that [itex]q<0[/itex] and then [itex]\vec{v}[/itex] is in the direction of P to S instead. You will get the same answer as far as all electromagnetism calculations are concerned, however now the angle would be [itex]90-\Theta[/itex]

    So what I'm saying is that the answer really could go either way. I was just trying to show you that the book is obviously having in mind the velocity of moving positive charge as a vector with which to measure the angle from the magnetic field.

    I didn't want to jump to this as a conclusion though because you didn't actually write out the question as it was written in the book, so perhaps the wording of the book is better. I was just hoping that by justifying the answer the book gave it would shed some light on it for you.
    But if that is really what the book asked then I would say it could truly be either answer.

    Don't beat yourself up over this anyway... This is a very artificial question. Introductory books quite often mess up on questions like this because they make assumptions without realizing it.. It is a source of some headache until you start getting into real electromagnetism calculations where you will see that you will be able to use either angle (provided you pick the correct corresponding charge and velocity) and you will be able to get the right answers.
  13. Aug 22, 2011 #12
    Thanks again, sir. Its a cliched word but what else word i have to describe your efforts in making me understand the concept.
    As u are referring velocity and conventional current, i interpreted in this way, that is
    1) q>0, for conventional current, similarly v(velocity) will be greater than 0, that is positive when i proceed from S to P, so since velocity vector is positive at the termination point of current, that is at P, i have to measure the angle from P.
    Am i sounding correct or silly?
  14. Aug 22, 2011 #13
    Yes that is absolutely correct.
    and that is why the book chose that direction.

    However, I also was pointing out that if you chose to use q<0 which would be the actual electrons then the velocity vector will be the other direction. (because the electrons will be moving in the opposite direction of conventional current)
    How do you flip a vector around? You multiply by -1 so the negative from the q and the negative from the change in direction cancel out and you get the same exact answers as if you used q>0 and velocity in the direction of conventional current.

    So both ways are correct... the difference is that when q<0 you are now measuring the angle the other way around.
    That is why I was saying that measuring it from either side is technically correct..

    But for future problems with that book, you should take it to be the direction of the conventional current with q>0 because that is clearly what they are assuming.

    So think about it this way..
    Measuring it from the point P (which is the direction in which the conventional current is traveling) is the "conventional way" to do it. There are other ways, but the book selected the "conventional" way as the "correct answer".
    Does that make you feel better about it?
  15. Aug 23, 2011 #14
    Now i feel so good after your enlightening thoughts. Nice way of putting things and u have mastered it. Thanks a ton sir.
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