Motion of a particle.

Question

I first started by integrating both sides with respect to t (dt). It says that B is along the z-axis but how do I account for that.

Answer 1

The question states that $\mathbf{B}$ is uniform, so this means that it is independent of time $t$, and independent of space $x,y,z$. So it must take the form \begin{equation} \mathbf{B} = B_1 \mathbf{i} + B_2 \mathbf{j} + B_3 \mathbf{k}, \end{equation} where $B_1, B_2, B_3$ are all constants. Now the question states that the $\mathbf{B}$ field is only in the $z$ direction, this means that the $x$ and $y$ components are zero, that is, $B_1 = B_2 = 0,\ B_3 \neq 0$.

The next step is to evaluate the cross product on the RHS, to do this you should write the velocity vector in a similar way to the magnetic field, i.e.

\begin{equation} \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}. \end{equation}

But now $v_1, v_2, v_3$ are not necessarily independent of time (and in fact $v_1, v_2$ do depend on time as you will see).

Now you can split this simplified ODE system into its individual components, there is one for each of the $\mathbf{i}, \mathbf{j}, \mathbf{k}$ directions. So you now have a system of three, first order, coupled ODES for the components of the velocity vector. These ODEs are quite easy to solve, you can then apply the initial conditions to the general solutions to produce the final result given in the problem.

Answer 2

Set $$\omega = \frac {qB}m$$ Then the components of $\mathbf v \times \mathbf B$ are such that the problem is the solution of the system $$\begin {cases} \dot{v}_x = \;\;\,\omega\, v_y \\ \\ \dot{v}_y = -\omega\, v_x \\ \\ \dot {v}_z = \quad\;\; 0 \end {cases}$$ with the given initial conditions.

Differentiating the first equation and using the second one, you obtain $$\ddot{v}_x = -\omega^2\, v_x$$ whose solution is $$v_x = v_0 \sin \omega t$$ From the first equation, you get $$v_y = v_0 \cos \omega t$$ Integrating you have $$x=\frac {v_0}{\omega}-\frac {v_0}{\omega} \cos \omega t$$ and $$y=\frac {v_0}{\omega} \sin \omega t$$ Note that $$\left(x - \frac {v_0}{\omega}\right)^2 + y^2 = \frac {v_0^2}{\omega^2}$$ The third equation gives $$z=0$$ so the path of the particle is a circle of center $$\left(\frac {mv_0}{qB},0\right)$$ on the xy-plane.

Finally $$v_x^2 + v_y^2 = v_0^2$$ so the motion is uniform with speed $v_0$ .