0
$\begingroup$

So my professor gave us the following question:

A particle with electric charge $Q$ and mass $M$ is initially traveling with velocity $v_0$ in the $x$ direction at time $t= 0$. There is a constant electric field of strength $E$ in the $y$ direction, and a constant magnetic field $B$ in the $z$ direction. To save a lot of ink, set $Q=M= 1$ while you work this problem.

a) Find the velocity $(v_x, v_y, v_z)$ as a function of time.

b) Find the position $(x, y, z)$ as a function of time assuming it started at $(0, 0, 0)$ at time $t=0$.

c) Rewrite your answer to part (b) with the $Q$ and $M$ factors properly included.

For context, we have been solving equations of motion by solving the differential equation $dv=dtF/m$ by separation of variables.

Here's my attempt at the situation:

a) The force experienced by the particle is the Lorenz force given by $F=q(E+(v \times B))$ . Since the electric field is given only in the y direction, and the magnetic field is given only in the z direction, and the initial velocity is given in the x direction, I deduced the following equations:

$F_x=V_0x \times B_z$

$F_y= E_y + V_0x \times B_z$

$F_z=V_0x \times B_z$

Problem is, I do not know how to go about this from here. Do I simply integrate the terms? And if so, how do I integrate a cross product? Or should I substitute expressions for $E$ and $B$ and then integrate? Even more importantly, are my equations even correct? I have a feeling that there is something wrong.

$\endgroup$
  • $\begingroup$ Why is the "v" upper case in your equations? If that is supposed to be the initial velocity, that's okay, but then these equations only satisfy the initial conditions. Try to make it more general by leaving the force in terms of $\mathbf{v}$ or $\dot{\mathbf{x}}$, which ever you prefer. $\endgroup$ – honeste_vivere Feb 2 '16 at 1:12
1
$\begingroup$

You've taken a wrong turn with the cross product business. A cross product is a vector operation - it maps two vectors to a third vector. However, you seem to be imagining that your three equations for $F_x$, $F_y$, and $F_z$ have a cross product in them, but they can't, because those equations contain only scalars.

The Lorentz equation as written like this

$$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$

contains a cross product between $\vec{v} = (v_x, v_y, v_z)$ and $\vec{B} = (0, 0, B_z)$. You can't take the cross product between $v_x$ and $B_z$, because those are scalar components of a vector, not vectors themselves.

What you want to do instead is separate that vector Lorentz equation into three scalar equations. To do that, you'll need to understand how to compute the cross product: $$\vec{v} \times \vec{B} = \left(\begin{matrix} v_y B_z - v_z B_y\\ v_z B_x - v_x B_z\\ v_x B_y - v_y B_x\\ \end{matrix}\right)$$

This can be simplified a lot, since we know that the magnetic field is entirely in the z-direction:

$$\vec{B} = \left(\begin{matrix} 0\\ 0\\ B_z\\ \end{matrix}\right)$$

so $B_x = 0$ and $B_y = 0$. Therefore, $$\vec{v} \times \vec{B} = \left(\begin{matrix} v_y B_z\\ - v_x B_z\\ 0\\ \end{matrix}\right)$$

Now we can create our full vector Lorentz equation:

$$ \left(\begin{matrix} F_x\\ F_y\\ F_z\\ \end{matrix}\right) = q \left[ \left(\begin{matrix} 0\\ E_y\\ 0\\ \end{matrix}\right) + \left(\begin{matrix} v_y B_z\\ - v_x B_z\\ 0\\ \end{matrix}\right) \right]$$

Now we can separate this equation into three equations, one for each component:

$$F_x = q v_y B_z$$ $$F_y = q (E_y - v_x B_z)$$ $$F_z = 0$$

Now, you can bring Newton's 2nd law in to actually create equations of motion; in vector form,

$$\left(\begin{matrix} F_x\\ F_y\\ F_z\\ \end{matrix}\right) = m \left(\begin{matrix} a_x\\ a_y\\ a_z\\ \end{matrix}\right) = m \frac{d}{dt}\left(\begin{matrix} v_x\\ v_y\\ v_z\\ \end{matrix}\right)$$

Putting that together with our Lorentz component equations,

$$m \frac{dv_x}{dt} = q v_y B_z$$ $$m \frac{dv_y}{dt} = q (E_y - v_x B_z)$$ $$m \frac{dv_z}{dt} = 0$$

This is a system of equations that, with some further algebraic and calculus manipulation, in combination with each other, can be integrated!


An addendum - the notation in your attempt is confusing - is $V_0x$ supposed to be the initial velocity in the x direction? If so, that should be the $v_x(t)$, not $v_{0x}$, since you want to solve for the motion over time, not just at the beginning. Once you've solved for the motion, you should use your initial velocity as an initial condition for the integration:

$$\vec{v}(t=0) = \left(\begin{matrix} v_0\\ 0\\ 0\\ \end{matrix}\right)$$


Note: I know you said to set the charge and the mass to unity, but it just seems like a waste of a perfectly good derivation to get to the end and not be able to tell how the charge and mass affect the motion. I hope the ink shortage is quickly alleviated wherever your professor lives.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.