We have seen that a moving charge develops its own magnetic field around it. Hence if a charge moves in a magnetic field the magnetic field due to the moving charge adds to the magnetic field in the surroundings and experiences a resultant force.
The magnitude of force depends on the magnetic field, speed of the charged particle and the charge of moving particle. Such a magnetic force experienced by a moving charge in a magnetic field is known as magnetic Lorrentz force. It is given by
F=q(vxB)
The direction of magnetic field is perpendicular to both the direction of motion and the direction of magnetic field and is given by Fleming's left hand rule.
Stretch the forefinger, middle finger and the thumb of left hand in a mutually perpendicular direction. If the fore finger represents the direction of magnetic field, middle finger represents the direction of motion then the thumb points in the direction of magnetic Lorentz force.
In a combined electric and magnetic field a moving charge experiences electric as well as magnetic force. Hence the total force experienced is the vector sum of electric and magnetic force. Thus the total Lorrentz force experienced by a moving charged particle is
F=Fe + Fm
F= qE + q(V x B)
Now let us discuss the Lorrentz force in different cases:
a) A charged particle moving parallel to the magnetic field
A charged particle moving parallel to a magnetic field will not experience any magnetic force and moves undeviated.
b) A charged particle moving perpendicular to the magnetic field
A charged particle entering a perpendicular magnetic field will get deflected continously and follows a circular path.
Illustration of a charged particle entering in a perpendicular uniform magnetic field
Let a particle of mass m and charge q enters a uniform magnetic field in a perpendicular direction with a speed v. In the following diagram the direction of magnetic field is into the plane of the paper and the particle enters from left to right at A.
In the above diagram, the direction of force at all the points A,B,C are towards the point O. It will be continuously deflected from its straight line path by a radial force qvB. Hence it follows a circular path. Now we shall evaluate the radius of the circular path followed by the particle.
The centripetal force required for the particle to move in path of radius r is
Fc=(mv2)/r
The magnitude of Lorentz force available is F=qvB.
Hence we can write
mv2/r=qvB
On solving above equation we get
r=mv/q/B= p/qB
where p is the momentum of the particle.
Thus the radius of path followed by charged particle depends on the momentum & magnitude of charge of the particle and magnitude of magnetic field.
The period of revolution of the particle
T=2pir/v =2pim/qB
And hence frequency of revolution is
f=qB/2pim
It should be noted that frequency is independent of speed.
The magnitude of force depends on the magnetic field, speed of the charged particle and the charge of moving particle. Such a magnetic force experienced by a moving charge in a magnetic field is known as magnetic Lorrentz force. It is given by
F=q(vxB)
The direction of magnetic field is perpendicular to both the direction of motion and the direction of magnetic field and is given by Fleming's left hand rule.
Stretch the forefinger, middle finger and the thumb of left hand in a mutually perpendicular direction. If the fore finger represents the direction of magnetic field, middle finger represents the direction of motion then the thumb points in the direction of magnetic Lorentz force.
In a combined electric and magnetic field a moving charge experiences electric as well as magnetic force. Hence the total force experienced is the vector sum of electric and magnetic force. Thus the total Lorrentz force experienced by a moving charged particle is
F=Fe + Fm
F= qE + q(V x B)
Now let us discuss the Lorrentz force in different cases:
a) A charged particle moving parallel to the magnetic field
A charged particle moving parallel to a magnetic field will not experience any magnetic force and moves undeviated.
b) A charged particle moving perpendicular to the magnetic field
A charged particle entering a perpendicular magnetic field will get deflected continously and follows a circular path.
Illustration of a charged particle entering in a perpendicular uniform magnetic field
Let a particle of mass m and charge q enters a uniform magnetic field in a perpendicular direction with a speed v. In the following diagram the direction of magnetic field is into the plane of the paper and the particle enters from left to right at A.
In the above diagram, the direction of force at all the points A,B,C are towards the point O. It will be continuously deflected from its straight line path by a radial force qvB. Hence it follows a circular path. Now we shall evaluate the radius of the circular path followed by the particle.
The centripetal force required for the particle to move in path of radius r is
Fc=(mv2)/r
The magnitude of Lorentz force available is F=qvB.
Hence we can write
mv2/r=qvB
On solving above equation we get
r=mv/q/B= p/qB
where p is the momentum of the particle.
Thus the radius of path followed by charged particle depends on the momentum & magnitude of charge of the particle and magnitude of magnetic field.
The period of revolution of the particle
T=2pir/v =2pim/qB
And hence frequency of revolution is
f=qB/2pim
It should be noted that frequency is independent of speed.
Work done by Lorentz force on a moving charge
We have learnt that a moving charge experiences a magnetic force magnetic field. As the particle is moving it has a displacement. Then what is the work done by the magnetic force in deflecting the particle?
As the direction of magnetic force is perpendicular to the direction of motion no work is done by the Lorentz force and hence the kinetic energy of particle will not change..
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It's a nice post. I really like it. Thanks for sharing. But if you provide some examples of Lorentz force. Then it'll be a great post.
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