Question

A beam of light travelling along x-axis is described by the electric field $E_y = (600 \frac{V}{m}) \sin \omega (t-\frac{x}{c})$.

The maximum magnitude of magnetic force on a charge $q = 2e$, moving along y-axis with the speed of $3 \times 10^2$ m/s is ($e = 1.6 \times 10^{-19}$ C)

Answer 1

1.92 × 10^{-22} N

Answer 2

The problem asks for the maximum magnitude of the magnetic force on a given charge moving in the presence of an electromagnetic wave.

1. Identify the maximum electric field strength ($E_0$): The electric field is given by $E_y = (600 \frac{V}{m}) \sin \omega (t-\frac{x}{c})$. From this equation, the maximum amplitude of the electric field is $E_0 = 600 \frac{V}{m}$.

2. Calculate the maximum magnetic field strength ($B_0$): For an electromagnetic wave, the magnitudes of the electric and magnetic fields are related by the speed of light $c$: $E_0 = B_0 c$ Given $c = 3 \times 10^8 \frac{m}{s}$ (speed of light in vacuum). $B_0 = \frac{E_0}{c} = \frac{600 \frac{V}{m}}{3 \times 10^8 \frac{m}{s}} = 2 \times 10^{-6} \text{ T}$.

3. Determine the direction of the magnetic field ($\vec{B}$): The light beam travels along the x-axis (direction of propagation). The electric field $\vec{E}$ is along the y-axis. In an electromagnetic wave, the direction of propagation ($\vec{k}$), the electric field ($\vec{E}$), and the magnetic field ($\vec{B}$) are mutually perpendicular, and their directions are related by $\vec{k} = \vec{E} \times \vec{B}$. Since the propagation is along +x ($\hat{i}$) and $\vec{E}$ is along +y ($\hat{j}$), then $\vec{B}$ must be along +z ($\hat{k}$), because $\hat{i} = \hat{j} \times \hat{k}$. So, the magnetic field $\vec{B}$ oscillates along the z-axis.

4. Calculate the maximum magnetic force ($F_{B,max}$): The charge $q = 2e = 2 \times 1.6 \times 10^{-19} \text{ C} = 3.2 \times 10^{-19} \text{ C}$. The charge moves along the y-axis with a speed $v = 3 \times 10^2 \frac{m}{s}$. So, the velocity vector $\vec{v}$ is along the y-axis. The magnetic force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by $\vec{F}_B = q(\vec{v} \times \vec{B})$. The magnitude of this force is $F_B = q v B \sin \theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Here, $\vec{v}$ is along the y-axis and $\vec{B}$ is along the z-axis. Thus, the angle between them is $\theta = 90^\circ$, which means $\sin \theta = 1$. The maximum magnetic force occurs when the magnetic field is at its maximum magnitude $B_0$. $F_{B,max} = q v B_0 \sin 90^\circ$ $F_{B,max} = (3.2 \times 10^{-19} \text{ C}) \times (3 \times 10^2 \frac{m}{s}) \times (2 \times 10^{-6} \text{ T}) \times 1$ $F_{B,max} = (3.2 \times 3 \times 2) \times (10^{-19} \times 10^2 \times 10^{-6}) \text{ N}$ $F_{B,max} = 19.2 \times 10^{(-19 + 2 - 6)} \text{ N}$ $F_{B,max} = 19.2 \times 10^{-23} \text{ N}$ $F_{B,max} = 1.92 \times 10^{-22} \text{ N}$

The final answer is $\boxed{1.92 \times 10^{-22} \text{ N}}$.

Explanation of the solution:

1. Obtain maximum electric field $E_0$ from the given equation.
2. Calculate maximum magnetic field $B_0$ using $B_0 = E_0/c$.
3. Determine the direction of the magnetic field (z-axis) based on the wave propagation (x-axis) and electric field direction (y-axis).
4. Calculate the maximum magnetic force using $F_{B,max} = qvB_0 \sin\theta$. Since velocity is along y-axis and magnetic field along z-axis, $\theta = 90^\circ$.

Answer: $1.92 \times 10^{-22} \text{ N}