Question

Floating in front of you there is a long, horizontal wire which carries a 3.50A current from East to West. At the same altitude as the wire and 0.50m from it, there is an electron which, at this moment, is also moving from East to West. The electron is between you and the wire. The kinetic energy of the electron is 4.56·10-19J. What is the size and direction of the force on the electron? Your answer should be some number of Newtons in a direction chosen from { North, South, East, West, Up, Down }.

Answer 1

First, we have to establish that the directions are placed on the cartesian plot as:

• North (y-axis, positive direction) and south (y-axis, negative direction)
• East (x-axis, positive direction) and west (x-axis, negative direction)
• Up (z-axis, positive direction) and down (z-axis, negative direction)

The force exerted on the electron can be found as $\overrightarrow{F_{elec/B}}=q_{elec}\overrightarrow{v_{elec}}\times\overrightarrow{B_{wire}}$ .

The velocity of the moving electron goes from east to west thus $\overrightarrow{v_{elec}} = - v_{electron}\widehat{i}$ . From the kinetic energy we can derive an expression for the instant velocity of the electron:

$E_{kinetic} = \frac{1}{2}m_{elec}[v_{elec}]^2 \\ \implies v_{elec} =\sqrt{\dfrac{2E_{kinetic}}{m_{elec}}}$

If we analyze the magnetic field around a long, straight, current-carrying conductor, it has axial symmetry about the x-axis (since the current goes from east to west). Hence must have the same magnitude at all points on a circle centered on the conductor and lying in a plane perpendicular to it (in this case the yz plane), and the direction of B must be everywhere tangent to such a circle (those directions can be determined by the right-hand rule as well). Thus, at all points on a circle of radius r around the conductor (r would be the distance between the electron and the wire), the magnitude B is

$B_{wire} = \dfrac{\mu_0I}{2\pi r}$ , the direction would be the z-axis (up direction) thus $\overrightarrow{B_{wire}} = B_{wire}\widehat{k}$ .

We use $\overrightarrow{F_{elec/B}}=q_{elec}\overrightarrow{v_{elec}}\times\overrightarrow{B_{wire}}$ to find the direction and the magnitude of the force on the electron:

$\overrightarrow{F_{elec/B}}=q_{elec}(-\sqrt{\frac{2E_{kinetic}}{m_{elec}}}\,\widehat{i})\times(\dfrac{\mu_0I}{2\pi r}\,\widehat{k}) \\ \overrightarrow{F_{elec/B}}= \dfrac{\mu_0Iq_{elec}}{\pi r}\sqrt{\frac{E_{kinetic}}{2m_{elec}}}(\,-\,\widehat{i}\times\widehat{k} \,) \\ \overrightarrow{F_{elec/B}}= \dfrac{\mu_0Iq_{elec}}{\pi r}\sqrt{\frac{E_{kinetic}}{2m_{elec}}} \begin{pmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ \overrightarrow{F_{elec/B}}= \dfrac{\mu_0Iq_{elec}}{\pi r}\sqrt{\frac{E_{kinetic}}{2m_{elec}}} \begin{pmatrix} -1 & 0 \\ 0 & 1 \\ \end{pmatrix}(-\widehat{j})$

After reduction and simplification we find $\overrightarrow{F_{elec/B}}= \dfrac{\mu_0Iq_{elec}}{\pi r}\sqrt{\frac{E_{kinetic}}{2m_{elec}}}\,\widehat{j}$ (which also means that the force goes into the y-axis positive direction that is north for our system), last substitution and the magnitude of the force will be:

$F_{elec/B}= \frac{( 4π × 10^{−7} N/A^2)(3.50\,A)(1.609 × 10^{−19} C)}{\pi (0.50\,m)}\sqrt{\frac{4.56×10^{−19}J}{2(9.109×10^{−31}kg)}}$

$F_{elec/B}=2.254 × 10^{−19} N; \overrightarrow{F_{elec/B}}= F_{elec/B}\,\widehat{j}$ .

In conclusion, we found the magnitude of the force as $F_{elec/B}=2.254 × 10^{−19} N$ and is directed to the north direction ($\overrightarrow{F_{elec/B}}= F_{elec/B}\,\widehat{j}$ ).

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.