Question

A charged particle is in a magnetic field within a three-dimensional space described by the standard Cartesian axes x, y, z. The magnetic field strength and direction varies with time and space. Initially, the particle has velocity $v_x = 200 \ m \ s^{-1}$ and $v_y = 210 \ m \ s^{-1}$. After travelling in the magnetic field for some time, the particle has travelled a distance of d = 500 m and has $v_z = 290 \ m \ s^{-1}$. Find $||$ (in $m/s^2$), the magnitude of the average acceleration of the particle. Assume there is no gravity.

Answer 1

237.86

Answer 2

The force on a charged particle in a magnetic field is given by $\vec{F} = q(\vec{v} \times \vec{B})$. The magnetic force is always perpendicular to the velocity vector ($\vec{F} \cdot \vec{v} = 0$). This means the magnetic force does no work on the particle.

According to the work-energy theorem, the change in kinetic energy of the particle is equal to the work done by the net force. Since there is no gravity and the magnetic force does no work, the net work done is zero.

$W_{net} = \Delta K = 0$.

This implies that the kinetic energy, and thus the speed, of the particle remains constant.

$v_i = v_f = v$.

The initial velocity components are given as $v_{ix} = 200 \ m/s$ and $v_{iy} = 210 \ m/s$. The problem statement "Initially, the particle has velocity $v_x = 200 \ m \ s^{-1}$ and $v_y = 210 \ m \ s^{-1}$" implies that the initial velocity vector is $\vec{v}_i = (200, 210, 0) \ m/s$.

The initial speed is $v_i = |\vec{v}_i| = \sqrt{v_{ix}^2 + v_{iy}^2 + v_{iz}^2} = \sqrt{200^2 + 210^2 + 0^2} = \sqrt{40000 + 44100} = \sqrt{84100} = 290 \ m/s$.

Since the speed is constant, the final speed is also $v_f = 290 \ m/s$.

The final velocity has a component $v_{fz} = 290 \ m/s$. Let the other components be $v_{fx}$ and $v_{fy}$.

The final speed is $v_f = |\vec{v}_f| = \sqrt{v_{fx}^2 + v_{fy}^2 + v_{fz}^2}$.

$290 = \sqrt{v_{fx}^2 + v_{fy}^2 + 290^2}$.

$290^2 = v_{fx}^2 + v_{fy}^2 + 290^2$.

This implies $v_{fx}^2 + v_{fy}^2 = 0$. Since squares of real numbers are non-negative, this is only possible if $v_{fx} = 0$ and $v_{fy} = 0$.

So, the final velocity vector is $\vec{v}_f = (0, 0, 290) \ m/s$.

The change in velocity is $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = (0 - 200, 0 - 210, 290 - 0) = (-200, -210, 290) \ m/s$.

The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{(-200)^2 + (-210)^2 + 290^2} = \sqrt{40000 + 44100 + 84100} = \sqrt{168200} \ m/s$.

$168200 = 1682 \times 100 = (2 \times 841) \times 100 = (2 \times 29^2) \times 10^2$.

$|\Delta \vec{v}| = \sqrt{2 \times 29^2 \times 10^2} = 29 \times 10 \times \sqrt{2} = 290 \sqrt{2} \ m/s$.

The distance travelled by the particle is given as $d = 500 \ m$. Since the speed is constant, the time taken to travel this distance is $\Delta t = \frac{d}{v}$.

$\Delta t = \frac{500 \ m}{290 \ m/s} = \frac{50}{29} \ s$.

The average acceleration is defined as $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$.

The magnitude of the average acceleration is $|| = \frac{|\Delta \vec{v}|}{\Delta t}$.

$|| = \frac{290 \sqrt{2} \ m/s}{\frac{50}{29} \ s} = \frac{290 \sqrt{2} \times 29}{50} \ m/s^2$.

$|| = \frac{29 \times 10 \times \sqrt{2} \times 29}{5 \times 10} = \frac{29 \times \sqrt{2} \times 29}{5} = \frac{29^2 \sqrt{2}}{5} = \frac{841 \sqrt{2}}{5} \ m/s^2$.

Calculating the numerical value:

$|| = \frac{841 \times \sqrt{2}}{5} \approx \frac{841 \times 1.41421356}{5} \approx \frac{1189.323}{5} \approx 237.8646 \ m/s^2$.