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    could some one please explain to me:

    the right hand thumb rule

    and fleming's left hand rule


    in a more simplified way

    thank u
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    The hand rule always gave me cramp as i tried every variation with my fingers and diretion. You also look like an arse doing it in your exam.

    Sorry cant help i was just remembering the good 'ole days of GCSE
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    Left hand rule...

    Point your thumb upwards. Point your first finger forwards. Point your second finger to the right. They should be at right angles.

    Now move your hand so the first finger points in the direction of the magnetic field (N to S). And your seecond finger points in the direction that the conventional current flows (+ to -). The direction of your thumb is the direction that the conductor with the current in will move.

    This is called the motor effect.

    Right hand rule...

    It's similar really. Your first finger should point in the direction of the field again, and your thumb in the direction of motion. Your second finger will show the direction of the current produced.

    This is called the dynamo effect.

    BTW Physics subforum anyone???
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    (Original post by thefish_uk)
    Left hand rule...

    Point your thumb upwards. Point your first finger forwards. Point your second finger to the right. They should be at right angles.

    Now move your hand so the first finger points in the direction of the magnetic field (N to S). And your seecond finger points in the direction that the conventional current flows (+ to -). The direction of your thumb is the direction that the conductor with the current in will move.

    This is called the motor effect.

    Right hand rule...

    It's similar really. Your first finger should point in the direction of the field again, and your thumb in the direction of motion. Your second finger will show the direction of the current produced.

    This is called the dynamo effect.

    BTW Physics subforum anyone???
    Yeah, that's it. Movement/motion is generated by force.

    Right hand rule is for inductive effect (dynamo effect is, I guess, the same).
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    Wiki Support Team
    The right hand thumb rule is different from the generator rule, and goes like this:

    Clench your right hand into a fist, with the thumb sticking out. Take a current-carrying wire, and stick your thumb in the direction that the current is flowing. The direction in which your fingers are curling (from base to tip) will tell you which direction the field lines produced by the wire go - either clockwise, or anticlockwise.

    My teacher always called it the "I'm enjoying physics" rule, which might help you to remember it (think thumbs-up or "Yo!" sign).
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    (Original post by pumakel)
    could some one please explain to me:

    the right hand thumb rule

    and fleming's left hand rule


    in a more simplified way

    thank u
    Whenever you have a cross product the following is the easiest way in my opinion.

    A cross B

    Point your right arm in teh direction of A, align your finger along the direction of B. Your thumb now points in the direction of A cross B.

    In teh case of an electric current generating a magnetic field, point your thumb in teh direction of teh current and thenyour finger points in the direction of the magnetic field. If you want to know the direction of the electric induced field from a change in magnetic flux, use lens's law in combination with the above.
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    thanks its a bit late but thnaks a lot we did the exam 3 weeks ago thanks for your intrest though it is much appreciated
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    (Original post by Jonatan)
    Whenever you have a cross product the following is the easiest way in my opinion.

    A cross B

    Point your right arm in teh direction of A, align your finger along the direction of B. Your thumb now points in the direction of A cross B.

    In teh case of an electric current generating a magnetic field, point your thumb in teh direction of teh current and thenyour finger points in the direction of the magnetic field. If you want to know the direction of the electric induced field from a change in magnetic flux, use lens's law in combination with the above.
    Our lecturers always used the 'corkscrew' method... :stupid:
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    (Original post by Mehh)
    Our lecturers always used the 'corkscrew' method... :stupid:
    Well I suppsoe if you , liek me, is a fan of red wine that may be an easy way to remember it :p:
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    (Original post by Jonatan)
    Well I suppsoe if you , liek me, is a fan of red wine that may be an easy way to remember it :p:
    But what if you have a left handed corkscrew. OHHHH hence left handed co-ordinate system
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    (Original post by Mehh)
    But what if you have a left handed corkscrew. OHHHH hence left handed co-ordinate system
    Heh, two words for you "linear algebra". I still don't understand the change of basis matrix fully I mean I know what it does, it is just teh algorithm to find it given two non-fundamental sets of basis vectors that is about as clear as a dirty windshield in a sandstorm to me.
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    (Original post by Jonatan)
    Heh, two words for you "linear algebra". I still don't understand the change of basis matrix fully I mean I know what it does, it is just teh algorithm to find it given two non-fundamental sets of basis vectors that is about as clear as a dirty windshield in a sandstorm to me.
    Do you mean the Jacobian?
    IE working out the determinate of the matrix where all the elements look like ∂x/∂θ ∂y/∂r ∂z/∂Φ all other variant combinations.

    If that is what you mean then its to work out the difference in volume (if we are talking about 3 orders of spacial freedom systems) between dxdydz at coordinate (x,y,z) and drdθdΦ [for instance] at the same coordinates. Hence you can simply do a variable change and of the x,y and z. Then replace dxdydz with JdrdΦdθ where J is the jacobian, then intergrate with respect to the new co-ordinate system, which is hopefully simplier to do.

    The matrix part of it is just a nice simple way to evaluate drdΦdθ/dxdydz at a range of points.
    Taking this into account I am sure you can prove the Jacobian is equivilent to taking the determinate of the particular matrix in question.

    BTW, what is a "fundamental" co-ordinate system?
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    (Original post by Mehh)
    Do you mean the Jacobian?
    IE working out the determinate of the matrix where all the elements look like ∂x/∂θ ∂y/∂r ∂z/∂Φ all other variant combinations.

    If that is what you mean then its to work out the difference in volume (if we are talking about 3 orders of spacial freedom systems) between dxdydz at coordinate (x,y,z) and drdθdΦ [for instance] at the same coordinates. Hence you can simply do a variable change and of the x,y and z. Then replace dxdydz with JdrdΦdθ where J is the jacobian, then intergrate with respect to the new co-ordinate system, which is hopefully simplier to do.

    The matrix part of it is just a nice simple way to evaluate drdΦdθ/dxdydz at a range of points.
    Taking this into account I am sure you can prove the Jacobian is equivilent to taking the determinate of the particular matrix in question.

    BTW, what is a "fundamental" co-ordinate system?
    Nope, Im talking about when you have a set of linearly independent basis vectors such as (0,1,0,0) , ( 1,0,0,1) , (1,0,0,0) , ( 0,0,1,0) and is then asked to find the matrix that transforms a general vector in this basis (a,b,c,d) into a vector expressed in another basis , such as (1,0,0,0), (0,1,0,0) , (0,0,-1,0) , (0,0,0,-1)
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    (Original post by Jonatan)
    Nope, Im talking about when you have a set of linearly independent basis vectors such as (0,1,0,0) , ( 1,0,0,1) , (1,0,0,0) , ( 0,0,1,0) and is then asked to find the matrix that transforms a general vector in this basis (a,b,c,d) into a vector expressed in another basis , such as (1,0,0,0), (0,1,0,0) , (0,0,-1,0) , (0,0,0,-1)
    You GOT to be kidding. You do realise that a matrix is simply a way of expressing a set of linear equations in one equation. Hence you just find the solution to a = α1 w + β1x+ γ1y + δz, the solution being the first line in the matrix.
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    (Original post by Mehh)
    You GOT to be kidding. You do realise that a matrix is simply a way of expressing a set of linear equations in one equation. Hence you just find the solution to a = α1 w + β1x+ γ1y + δz, the solution being the first line in the matrix.
    Oh yea, well now try to do it for teh exampel I mentioned and then ask yourself if you did not do it the wrongw ay arround. Then try to figure out what is really going on. Remember that the vector v = (1,2,3,4) refers to the combination of basis vectors a,b,c,d such that v = 1a + 2b + 3c +4d

    The question is now if the matrix that gives v in terms of the basis e,f,g,h should be composed as the set of linear equations expressing the basis a,b,c,d as fucntions of e,f,g,h or if it should be the set of equations expressing e,f,g,h in terms of a,b,c,d. In eitehr case proove the result rigorously.
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    I am assuming what you mean is that
    (a)(0 1 0 0) = (w)
    (b)(1 0 0 1) = (x)
    (c)(1 0 0 0) = (y)
    (d)(0 0 1 0) = (z)

    and

    (a')(1 0 0 0 ) = (w)
    (b')(0 1 0 0 ) = (x)
    (c')(0 0 -1 0) = (y)
    (d')(0 0 0 -1) = (z)

    So the two lines are equivilent

    (a)(0 1 0 0) = (a') (1 0 0 0 )
    (b)(1 0 0 1) = (b') (0 1 0 0 )
    (c)(1 0 0 0) = (c') (0 0 -1 0)
    (d)(0 0 1 0) = (d') (0 0 0 -1)

    Knowing that multiplication in the field of Matrices (specifically non singluar matrices) is a 'Group'. You can apply all that malarkey about groups to it and come up with the below.

    (a') = (a)(0 1 0 0)(1 0 0 0 )-1
    (b') = (b)(1 0 0 1)(0 1 0 0 )
    (c') = (c)(1 0 0 0)(0 0 -1 0)
    (d') = (d)(0 0 1 0)(0 0 0 -1)

    The matrix inversion and multiplication are left as an exercise for the reader.

    PS its damn hard to get the matrix to render properly so i can't be arsed

    BTW I am converting both sets of weird vector spaces into cartisian i j k stylie vectors using a matrix on each. Then inverting one of them allows me to convert to cartisian then unto the OTHER vector space.
    At the end of the day you're only playing with linear equations.
 
 
 
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