E and B Fields (Part 2)

Electricity and Magnetism Level 1

At time $t = 0$ , a particle of mass $m = 1$ and charge $q = +1$ is at rest at the origin in the $xyz$ coordinate system. There are uniform electric and magnetic fields $(\vec{E}$ and $\vec{B})$ throughout all of space.

$\vec{E} = (E_x, E_y, E_z) = (1,1,1) \\ \vec{B} = (B_x, B_y, B_z) = (1,1,1)$

If the spatial coordinates of the particle at time $t = 1$ are $(x_f, y_f, z_f)$ , enter your answer as $x_f + y_f + z_f$

The answer is 1.5.

1 solution

Karan Chatrath
Sep 2, 2019

Nice problem. This has a very elegant closed-form solution.

Consider the particle at a general instant of time at a general point $(x,y,z)$ . The force acting on this particle is:

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

Substituting all expressions and simplifying gives:

$\ddot{x} = \dot{y} - \dot{z} + 1$ $\ddot{y} = \dot{z} - \dot{x} + 1$ $\ddot{z} = \dot{x} - \dot{y} + 1$

Adding these three equations gives:

$\ddot{x} +\ddot{y} +\ddot{z} = 3$

Integrating the above expression gives:

$\dot{x} +\dot{y} +\dot{z} = 3t + c$

Here, by applying the initial conditions leads to:

$\dot{x} +\dot{y} +\dot{z} = 3t$

Integrating once again and applying initial conditions gives:

$\boxed{x + y+ z = \frac{3t^2}{2}}$

Substituting $t=1$ gives the required answer of $\boxed{1.5}$ .

Nice solution, thanks. I first found this out experimentally, and then was able to justify it as follows: During the first little $\Delta t$ , the electric field gives an impulse in the direction of $(1,1,1)$ . Then the velocity is perfectly aligned with the magnetic field, rendering the magnetic field irrelevant from then on.

Steven Chase - 1 year, 9 months ago

This is an intuitive perspective.

Karan Chatrath - 1 year, 9 months ago

E and B Fields (Part 2)

The answer is 1.5.

1 solution

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