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Magnetism
i can make ur concepts shake
None
rod length L
movin wid velocity V
magnetic field perpendicular 2 d plane of d paper
now it is said that a force due 2 magnetic field acts on d e- inside it due 2 which d e-s moves 2 one end of d rod
here d angle between V & B is 90 therefore d e- should move in circular path
thus it should not reach d end then how is d polarity created
Comments (8)
dinesh arora
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Joined: 26 Feb 2007
Posts: 9
9 Jan 2008 00:27:35 IST
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Well, there r two components of velocity, one is wasted the other is used to take the electrons to the corners.
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9 Jan 2008 02:51:06 IST
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I think ur concepts themselves are quite shaken up.
See do u know why electron follows circular path? It is becoz, when it is free to move and is moving perpendicular to mag field, its velocity keeps changing every moment and it keeps getting a push perpendicular to its direction and hence it moves in a circular path.
Here in the conductor, the electrons can't come out of it. Hence they are not free to move. They can move only along the length of conductor.
Now if U see in the figure, the mag field is inside the plane of the paper. Now if U apply right hand rule, U will find the velocity of proton is coming out to be upwards and hence eletron's velocity is downwards. Now as the electron is not free to move outside the conductor, its velocity can't keep changing continuosly and hence it cant follow circular path and so it follows a linear path along the conductor and creates a polarity at the ends.
Hope U understood.
9 Jan 2008 16:04:20 IST
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Hello,
Dude you have not mentioned the direction of velocity wrt the length of rod.
EMF would be induced only if v is perpendicular to length of the rod
If v is perpendicular to length then:
F = q(v X B)
Thus the magnetic force will be perpendicular to both v as well as B which is along the length of the rod. So where is the circular motion.???????
Electrons will move along the length of rod and constitute an emf...
Dude you have not mentioned the direction of velocity wrt the length of rod.
EMF would be induced only if v is perpendicular to length of the rod
If v is perpendicular to length then:
F = q(v X B)
Thus the magnetic force will be perpendicular to both v as well as B which is along the length of the rod. So where is the circular motion.???????
Electrons will move along the length of rod and constitute an emf...
9 Jan 2008 16:06:29 IST
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Always remember that emf is produced only when the conductor cuts the magnetic field lines.
Now when a rod moves in a magnetic field, the electrons and protons experience a force coz they are also moving with the rod. Any charge moving in a magnetic field experiences a force of magnetic origin.
for inducing magnetic field, the conductor must travel in a plane that is perpendicular to the magnetic field.
This can be veified by the fact that F = q(v x B).
Only electron will move in the direction of force coz protons are fixed at their positions.
Thus only if v is perp. to B the force is exerted on electrons. (Even if its not perpendicular only the perpendicular component will exert a force)
This can be verified from the expression that
induced emf = B (l x v)
Thus l must also be perpendicular to velocity. Even if its not, only the perp. component will affect it.
So, Also the direction of travel must be perpendicular to the length of rod.
Hence we get the required direction
In short, the rod must cut the magnetic field lines.....
Now when a rod moves in a magnetic field, the electrons and protons experience a force coz they are also moving with the rod. Any charge moving in a magnetic field experiences a force of magnetic origin.
for inducing magnetic field, the conductor must travel in a plane that is perpendicular to the magnetic field.
This can be veified by the fact that F = q(v x B).
Only electron will move in the direction of force coz protons are fixed at their positions.
Thus only if v is perp. to B the force is exerted on electrons. (Even if its not perpendicular only the perpendicular component will exert a force)
This can be verified from the expression that
induced emf = B (l x v)
Thus l must also be perpendicular to velocity. Even if its not, only the perp. component will affect it.
So, Also the direction of travel must be perpendicular to the length of rod.
Hence we get the required direction
In short, the rod must cut the magnetic field lines.....
15 Jan 2008 10:28:36 IST
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Here if the electron is moving then it will move in such a way that it wont touch the boundary (as radius<< bcz mv/bq; m<<<<), so one will think that it will move in a circle but actually at any instant it would have a horizontal vel. Due to which it will constantly move up.
And similarly proton will move down.and d polarity would be created.
15 Jan 2008 22:32:53 IST
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giving d details 4 d question
rod moving wid velocity v perpendicular 2 its length
direction of magnetic field is perpendicular 2 d plane of motion of d rod
r=mv/qB
that is d radius of d circle in which d e should move
as here v and B are perpendicular 2 each other.
(from d equation radius is coming very very small as compared 2 d thickness of d rod)











