Electromagnetic Field Parameters

The particle has mass $m = 9.0 \times 10^{-5} \text{ kg}$ and charge $q = 3.0 \times 10^{-4} \text{ C}$.

The uniform electromagnetic field is given by $\vec{E} = 2\hat{x} \text{ V.m}^{-1}$ and $\vec{B} = 3\hat{z} \text{ V.m}^{-2}\text{.s}$. Let $E_0 = 2$ and $B_0 = 3$.

The particle is released from rest at coordinates $\vec{r}(0) = (0, 5\text{ m}, 0)$ at time $t=0$. This starting position is on the y-axis.

Particle Motion Analysis

The equation of motion for a charged particle in an electromagnetic field is determined by the Lorentz force:

$ m\frac{d\vec{v}}{dt} = q(\vec{E} + \vec{v} \times \vec{B}) $

Substituting the given fields $\vec{E} = E_0 \hat{x}$ and $\vec{B} = B_0 \hat{z}$, we obtain the component equations of motion:

• $m\frac{dv_x}{dt} = q(E_0 + v_y B_0)$
• $m\frac{dv_y}{dt} = -q v_x B_0$
• $m\frac{dv_z}{dt} = 0$

Since the particle starts from rest ($v_x(0)=0$, $v_y(0)=0$, $v_z(0)=0$), $v_z(t)=0$ for all time. Solving the coupled differential equations for $v_x$ and $v_y$ yields:

$ v_x(t) = \frac{E_0}{B_0} \sin(\omega t) $

$ v_y(t) = \frac{E_0}{B_0} (\cos(\omega t) - 1) $

where $\omega = \frac{qB_0}{m}$ is the characteristic angular frequency.

Calculating Time to Return to Y-Axis

Integrating the velocity components to find the position, starting from $\vec{r}(0)=(0, 5, 0)$:

$ x(t) = \int v_x(t) dt = \frac{E_0}{B_0\omega} (1 - \cos(\omega t)) $

The $x$-position is zero at $t=0$ since the particle starts on the y-axis ($x(0)=0$). For $t>0$, the particle's $x$-velocity $v_x(t) = \frac{E_0}{B_0} \sin(\omega t)$ becomes non-zero, causing it to leave the y-axis. The particle returns to the y-axis when $x(t)=0$ again.

$ \frac{E_0}{B_0\omega} (1 - \cos(\omega t)) = 0 $

This condition implies $1 - \cos(\omega t) = 0$, or $\cos(\omega t) = 1$. The general solutions for $\omega t$ are $2n\pi$ where $n$ is an integer.

The possible times are $t = 0, \frac{2\pi}{\omega}, \frac{4\pi}{\omega}, \dots$. The question asks for the *first time* the particle *comes back* to the y-axis after $t=0$. This corresponds to the first non-zero solution:

$ t = \frac{2\pi}{\omega} $

Numerical Calculation

First, calculate the angular frequency $\omega$:

$ \omega = \frac{qB_0}{m} = \frac{(3.0 \times 10^{-4} \text{ C}) \times (3 \text{ T})}{9.0 \times 10^{-5} \text{ kg}} = \frac{9.0 \times 10^{-4}}{9.0 \times 10^{-5}} \text{ s}^{-1} = 10 \text{ s}^{-1} $

Now, calculate the time $t$:

$ t = \frac{2\pi}{\omega} = \frac{2\pi}{10 \text{ s}^{-1}} = \frac{\pi}{5} \text{ s} $

Approximating the value:

$ t \approx \frac{3.14159}{5} \approx 0.628318 \text{ s} $

Rounded off to two decimal places, the time $t$ is approximately $0.63$ seconds.