Motion of a charged particle in a magnetic field

In summary: I appreciate your help.In summary, the conversation discusses the ratio of angular frequencies of two identical charged particles entering a region of uniform magnetic field at different angles. The solution involves equating the Lorentz force and centripetal force equations and factoring in the component of velocity perpendicular to the magnetic field. The final result shows that the angular frequency is independent of the particles' velocities and the ratio is 1:1, which is confirmed by the book. The conversation also addresses a possible mistake in the initial solution and provides a corrected method.
  • #1
kihr
102
0

Homework Statement



Two identical charged particles moving with the same speed enter a region of uniform magnetic field. If one of these enters normal to the field direction, and the other enters at an angle of 30 degrees with the field, what would be the ratio of their angular frequencies?


Homework Equations



Lorentz force on the charged particle = qvBsin(theta) where theta is the angle between v and B.
Centripetal force on the particle as it moves along a circular path = mv^2/r



The Attempt at a Solution



Equating the above two equations we get

v=qBr / msin(theta)

Therefore w(omega)=qB/msin(theta) (since v=rw)

On the above basis the ratio of the angular frequencies is 1:2 (since theta is 90 deg. in one case, and 30 deg. in the other case).

Since the answer as quoted in books is 1:1, I am unable to find out where I have gone wrong. Would appreciate some tips in case my method is not correct. Thanks.
 
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  • #2
I don't think that you did wrong. Maybe there is typo in your book.
 
  • #3
You did make a mistake. You said the velocity of the particle is given by v*sin(theta), which is the component of the velocity perpendicular to the magnetic field B. This is the component of the velocity responsible for uniform circular motion in a magnetic field. However, you go on to say that v = rw, which is true in general, but for the current problem you must have the perpendicular velocity vsin(theta) = rw; the angular frequency is associated only with the component of velocity perpendicular to B.

The angular frequency is given by w = |q|B/m. The period T, frequency f, and angular frequency w, are independent of the speed of the particle, and hence, particles with the same charge-to-mass ratio have the same T, f, and w. They vary in their radii. Your book is correct.
 
  • #4
I have factored in the component of v normal to B, i.e. vsin(theta) in the solution. When I wrote v=rw, I quoted the general formula. You could work it out the way you have suggested, and the answer would still be identical to what I have got. Thanks.
 
  • #5
Let's work it out and find out.

The centripetal force is equivalent to the magnetic force. So we have

[tex] F_{cent}=\frac{m\left(vsin\theta\right)^{2}}{r} = \left|q\right|\left(vsin\theta\right)B = F_{B}[/tex]

where [tex] vsin\theta [/tex] is the component of velocity perpendicular to B. We can rewrite this as

[tex]vsin\theta= \frac{\left|q\right|rB}{m}[/tex]

Now, in general [tex] v = \omega r [/tex]. For this problem, it is only the component of velocity perpendicular to B that contributes to the angular frequency. If there is a component of velocity parallel to B, then the particle will have a helical trajectory. Taking these facts into account,

[tex] vsin\theta = \omega r [/tex]

(this looks like where you made your mistake) and substitution into the previous equation gives the correct result:

[tex]\omega = \frac{\left|q\right|B}{m}[/tex]

Hence, the angular frequency is independent of the particle's velocity. The ratio is 1:1 and your text is correct.
 
  • #6
Many thanks for giving me the clue. I have now understood where I had gone wrong.
 

1. How does a magnetic field affect the motion of a charged particle?

A magnetic field can exert a force on a charged particle, causing it to change direction and/or speed. This force is perpendicular to both the direction of the particle's motion and the direction of the magnetic field.

2. What is the equation for the force on a charged particle in a magnetic field?

The equation for the force on a charged particle (F) in a magnetic field is F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.

3. How does the mass of the charged particle affect its motion in a magnetic field?

The mass of a charged particle does not directly affect its motion in a magnetic field. However, it does affect the particle's velocity, which in turn affects the magnitude of the force exerted on it by the magnetic field.

4. Can a charged particle's motion in a magnetic field be predicted?

Yes, the motion of a charged particle in a magnetic field can be predicted using the Lorentz force law, which takes into account the charge, velocity, and magnetic field strength of the particle. However, other factors such as external forces and the shape of the magnetic field can also play a role in the particle's motion.

5. Does the direction of the magnetic field affect the direction of the force on a charged particle?

Yes, the direction of the magnetic field does affect the direction of the force on a charged particle. The force will always be perpendicular to both the direction of the particle's motion and the direction of the magnetic field, following the right-hand rule.

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