How does a charged particle move in Earth's magnetic field?

[Phyrrus]

Homework Statement

Assume the Earth's magnetic field is almost homogeneous with direction along the z-axis, with a small inhomogeneous modification which make the field lines converge towards the z-axis. Also ignore relativistic and gravitational effects.
.
First assume the magnetic field, B = B_{0} = B_{0}k, to be time independent and homogeneous, with
k as a unit vector in the z-direction. A particle with charge q and mass m is moving in this field.
Initially, at time t = 0 the particle has velocity v_{0}, with u_{0} as the component in the z-direction and
w0 as the component in the x; y-plane.
a) Write the vector form of the equation of motion of the particle and show that it has solutions of
the form
r(t) = ρ_{0}(cos \omega_{0}ti + sin \omega_{0}0tj) + v_{z}t k
Determine the constants ρ_{0}, ω_{0} and v_{0} in terms of the initial velocity and magnetic field strength B0.

Homework Equations

F=q(E+v\timesB)

The Attempt at a Solution

E=0
a=(q/m)(v_{0}\timesB_{0})

v_{0}=(w_{0},u_{0})
where w_{0}=|w_{0}|cos\varthetai+w_{0}sin\varthetaj)

therefore v_{0}\timesB_{0} = B_{0}|w_{0}|sin\varthetai-B_{0}w_{0}cos\varthetaj

 

[mfb]

What is ϑ?
For (a), all you have to do is to calculate the acceleration vector based on the given formula for r(t), and show that it satisfies the equation for the Lorentz force.

 

[Phyrrus]

The \vartheta was the angle used to separate w_{0} into x and y components.

But I'm sorry, can you please elaborate more? Do you mean all I need to do is differentiate r(t) twice and equate coefficients?

 

[mfb]

"show that it has solutions of the form" -> show that [equation] is a solution -> show that [equation] has the Lorentz force as acceleration.

Do you mean all I need to do is differentiate r(t) twice and equate coefficients?

With the calculated v0 x B0, right.

 

[Phyrrus]

Ok, but how do I calculate v_{0}\timesB_{0} when v_{0}=(w_{0},u_{0}) ? Don't I need to separate w_{0} into x and y components?

 

[mfb]

I would convert everything to polar coordinates, but splitting them in components in Cartesian coordinates is possible, too.

 

[Phyrrus]

I'm sorry, I still don't quite get it

 

[mfb]

Do you mean all I need to do is differentiate r(t) twice and equate coefficients?

Did you do that? What did you get?

 

[Phyrrus]

I'm not going to lie, I couldn't do it.

 

[mfb]

What did you get? Where did you run into problems?
Those are obvious follow-up questions, you could have answered them without an extra post from me...

 

[Phyrrus]

I'm still not entirely sure what I am supposed to do. Is this right?

d^2r/dt^2 = -\omega_{0}^{2}\rho_{0}(cos\omega_{0}t i + sin\omega_{0} j)

Now we can say that, the above acceleration expression is equal to the Lorentz acceleration of (q/m)*(v_{0}\timesB_{0}?

But now how do I split v_{0}into i,j,k components? I tried splitting it into components before with the angle theta, but that was obviously wrong.

 

[mfb]

$$\vec{v} \times \vec{B} = \begin{pmatrix}
v_j B_0 \\
-v_i B_0 \\
0
\end{pmatrix}$$ where I used that the B-field has a component in k-direction only.
v_i and v_j are just the i- and j-component of $\frac{d\vec{r}}{dt}$

 

[Phyrrus]

Ok, so then v_{j}=cos(\omega_{0}t) and -v_{i}=sin(\omega_{0}t)

And then the constants -\omega_{0}^{2}\rho_{0}=(q/m)*B_{0}

 

[mfb]

Ok, so then v_{j}=cos(\omega_{0}t) and -v_{i}=sin(\omega_{0}t)

I think there are prefactors ρ_0 ω_0 missing, but the concept is right.

 

[Phyrrus]

So once I find the prefactors, I am given a 3 set of relations from which I can solve for the 3 constants?

 

[mfb]

Right.

 

[Phyrrus]

Ok, thanks for all your help. How do I find them?

 

[mfb]

Find what?
It is all in the thread now, you just have to combine it.