Motion of a charged particle in a magnetic field

[kihr]

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.

 

[ApexOfDE]

I don't think that you did wrong. Maybe there is typo in your book.

 

[buffordboy23]

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.

 

[kihr]

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.

 

[buffordboy23]

Let's work it out and find out.

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

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

where $$vsin\theta$$ is the component of velocity perpendicular to B. We can rewrite this as

$$vsin\theta= \frac{\left|q\right|rB}{m}$$

Now, in general $$v = \omega r$$. 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,

$$vsin\theta = \omega r$$

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

$$\omega = \frac{\left|q\right|B}{m}$$

Hence, the angular frequency is independent of the particle's velocity. The ratio is 1:1 and your text is correct.

 

[kihr]

Many thanks for giving me the clue. I have now understood where I had gone wrong.