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Magnetic field

  1. Jul 4, 2012 #1
    So im kind of a newb to Physics, I am currently studying magnetic field and I had a question to ask about this topic:
    1. I know theres a formula to find the magnitude of the force, but I don't quite understand why the force is directly proportional to the angle and current. Is it because that if theres more current, theres more potential difference so theres more force?
     
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  3. Jul 5, 2012 #2

    Simon Bridge

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    Welcome to PF.
    You have to be more precise than that - there are a lot of magnetic force equations. It sounds like you are thinking of the force on a current-carrying loop by a magnetic field... is this correct?

    But basically the answer to "why questions is as follows: "them's the rules". In science we leave the why questions to philosophers and concentrate on questions we can conduct experiments to answer like "how" and what". The various rules tell you how the current and the magnetic field relates to the resultant motion through a force.

    I can go into details when know the situation that has puzzled you.
     
  4. Jul 5, 2012 #3

    vanhees71

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    It's easier to understand if you think about a single charged particle in a magnetic field first. It turns out (and can be proven from the action principle for a system of charged particles and electromagnetic fields) that the force on the particle (in non-relativistic approximation) in a magnetic field is given by

    [tex]\vec{F}=q \vec{v} \times \vec{B}.[/tex]

    Thus, the magnetic force on the particle acts always perpendicular to its velocity, and the above expression has to be evaluated at the actual time and position of the particle (locality of the electromagnetic interaction). This law is named Lorentz force after the great Dutch physicist H. A. Lorentz, who is most famous for the development of the classical theory of charged point particles (at his time he had electrons in mind which have been discovered by J. J. Thomson in 1897 as the first elementary particle ever).

    Now look at a continuous current. It is described by the current density [itex]\vec{j}[/itex], which gives by definition the charge per time, flowing through a unit area perpendicular to the current vector. It is given by

    [tex]\vec{j}(t,\vec{x})=\rho(t,\vec{x}) \vec{v}(t,\vec{x}).[/tex]

    Here, [itex]\rho[/itex] is the charge density of the flowing medium and [itex]\vec{v}[/itex] the velocity of the charges at the given space-time point.

    In a little volume [itex]\mathrm{d} V[/itex] around this point, the charge contained in it is given by

    [tex]\mathrm{d} Q=\mathrm{d} V \rho[/tex]

    and the Lorentz force on this little charge is

    [tex]\mathrm{d} \vec{F}=\mathrm{d} Q \vec{v} \times \vec{B}=\mathrm{d} V \rho \vec{v} \times \vec{B}=\mathrm{d} V \vec{j} \times \vec{B}.[/tex]

    This means that the force per volume element of the flowing fluid is given by

    [tex]\vec{f}=\vec{j} \times \vec{B}.[/tex]

    The most simple application is to a current through a thin wire. In this case you can assume the current constant across the cross section of the wire. Now you choose an arbitrary orientation for the direction of the wire. Let the tangent vector along the wire denoted by [itex]\mathrm{d} \vec{l}=\vec{n} \mathrm{d} l[/itex] (where [itex]\mathrm{d} l[/itex] is the length of a little piece of the wire and [itex]\vec{n}[/itex] the tangent unit vector along the wire at this point) and the total current [itex]I[/itex] by definition has the sign relative to this tangent vector, i.e., if the current runs in direction of the tangent vector its counted positive, otherwise negative. Thus we have

    [tex]\vec{j}=\frac{I}{A} \vec{n},[/tex]

    along the wire, where [itex]A[/itex] is the cross-sectional area of the wire. Thus according to the above given ideas, we have for the force on the little piece of the wire

    [tex]\mathrm{d} F=\underbrace{A \mathrm{d} l}_{\mathrm{d V}} \vec{j} \times \vec{B} = \mathrm{d} l I \vec{n} \times \vec{B}.[/tex]

    The total force acting on the wire is thus given by

    [tex]\vec{F}=\int_{\text{wire}} \mathrm{d} \vec{l} \times \vec{B}.[/tex]

    The magnitude of the force increment is

    [tex]|\mathrm{d} \vec{F}| = \mathrm{d} l |I| |\vec{B}| \sin[\angle (\vec{n},\vec{B})].[/tex]

    I hope this helps you to understand the concepts behind the Lorentz force better.
     
  5. Jul 11, 2012 #4
    Seeing that i have similar question to magnetic field, how do you calculate the magnetic strength of a magnet? In my text book which is crap only shows me F=qvB or F=nIlB( no. coils*current*length*MF strength)

    And another question i think my postulate is right, do magnetic field behave similarly to gravitational field except it doesn't bend spacetime, it has infinite range like gravity.
    I searched up B, apparently it is given by B=μIn. or B=μI/(2∏r)
     
  6. Jul 11, 2012 #5

    Simon Bridge

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    For a solid bar magnet, it depends on material properties so you just have to look it up. For an electromagnet, you sum the magnetic contributions from the current at each part of the wire to get the B field.

    But if you mean to calculate the strength of a magnet ... (as opposed to the stregth of the magnetic field.)
    http://spinalbrace.com/tectonic/strength.htm

    EM fields have a very long range, this is true. The range of a field depends on the mass of the particle that mediates it ... the more massive the particle, the shorter range the field. EM fields are mediated by photons: mass = 0. Score!

    It is not usually a good idea to think of them as being "like gravity" though. The similarities in the math are down to geometry.
     
  7. Jul 11, 2012 #6
    There is an unique property of light that many of my highschool cohort overlook is that it is bent by gravity? At highschool level the only formula we touch is the gravity formula which revolves around something with mass. Whereas light have no mass but it is still bent! This is shown in solar eclipse where we normally dont see stars behind the sun because the sun is too bright and covering its path, but in a solar eclipse it is bent by the Sun's massive gravity allowing us to see them.

    Although gravity affects matter with "mass" we know that light do have "mass" because of Mass-Energy equivalence, light is the transfer of energy with no matter, so it does get bent by gravity, however what i am really asking is what is the formula to calculate the amount in degrees that light will get bent?

    Though i haven't covered up to light yet however i believe when my class looks at the "paper and hole dot"experiment will all believe that light travels in straight line through the same medium however this may not be true?
     
    Last edited: Jul 11, 2012
  8. Jul 12, 2012 #7

    Simon Bridge

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    That's a very good and important observation even if it is not quite relevent to the original question.

    Light travels the quickest path between two points. Most of the time this is the same as the shortest path (a straight line) but not always. You know that light is bent when it crosses between two media for example ... the bend (that results in Snell's law) minimized the time of flight.

    To give you an idea of the difference - the shortest distance between the North and South pole of the Earth is right through the middle ... which would involve tunneling ... but the quickest path would be along the outside ... which is curved.

    This is what happens with light - gravity curves space-time so the quickest path becomes a curve. You don't need to have mass to be forced into this path. If you do have mass, you will be forced to take a slower path, that's the difference.

    Space is also curved in the classroom ... however, the distances you will be doing your experiments over are too small for your rulers to be able to measure the gravitational bending of light.

    You will also be taught that light travels in "rays" too.

    What I want you to draw from all this is as follows:
    It's all good - these are models of reality, not reality itself. Think of the physics you learn as a kind of map. Different maps have different uses.
     
    Last edited: Jul 12, 2012
  9. Jul 12, 2012 #8
    I loled when i saw this last sentence because i was watching videos about relativity to reinforce what i have learnt already, and in the video it said something similar that, there is no 1 correct map it is just the perspective is different relative to another map.

    So when we see the sun and stars do we see the real position or actual position. i have been recently watching videos from Ozmoroid on relativity and in his 3rd video "Relativity 3" he showed me some formulas on the bend of light could you explain them to me please Simon Bridge?
     
  10. Jul 12, 2012 #9

    Simon Bridge

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    :) In relativity the phrase "real position" is hard to define.
    Not seen the videos so I cannot explain them - probably best to ask the question in a different thread.
     
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