Question

An electron moves in a straight line at a speed of \(5.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\). What are the magnitude and the direction of the magnetic field created by the moving electron at a distance \(d=5.00 \mathrm{~m}\) ahead of it on its line of motion? How does the answer change if the moving particle is a proton?

Short answer

Answer: At a distance of 5.00 m ahead of both the electron and the proton on their respective line of motion, there is no magnetic field created by either particle. The magnitude of the magnetic field at this specific point is zero.

Step-by-step solution

Determine the magnetic field due to a moving charged particle

The magnetic field created by a moving charged particle like an electron or a proton at a certain point can be found using the Biot-Savart Law. The Biot-Savart Law, in this case, can be written as: $$ \vec{B}=\frac{\mu_{0}}{4\pi} \frac{q\vec{v} \times \vec{r}}{r^{3}} $$ Where: - \(\vec{B}\) is the magnetic field, - \(\mu_{0}\) is the vacuum permeability (\(4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A}\)), - \(q\) is the charge of the particle, - \(\vec{v}\) is the velocity of the particle, - \(\vec{r}\) is the position vector from the particle to the point where we calculate the magnetic field, and - \(r\) is the magnitude of \(\vec{r}\).

Calculate the magnetic field for the electron

We know that the distance \(d=5.00\,\mathrm{m}\) is along the line of motion in the same direction as the electron's velocity. We can write the position vector \(\vec{r}\) as: $$ \vec{r} = d\hat{v} $$ Where \(\hat{v}\) is the unit vector in the direction of \(\vec{v}\). Therefore, \(\vec{v} \times \vec{r}\) becomes zero as they are parallel, and the magnetic field at the point will be zero: $$ \vec{B_e} = 0 $$ So, there is no magnetic field created by the electron at the specific point ahead of it on its line of motion.

Calculate the magnetic field for the proton

Now we need to find the magnetic field created by a proton at the same point. Since the only difference between an electron and a proton is their charge (and mass, but the mass doesn't affect the magnetic field), we can use the same expression as in Step 1. The charge of the proton \(q_{p}\) is the same as the magnitude of the charge of the electron, but with a positive sign (\(1.60 \times 10^{-19} \mathrm{C}\)). However, since \(\vec{v} \times \vec{r}\) are parallel for this point, the magnetic field for the proton at this point will also be zero: $$ \vec{B_p} = 0 $$

Compare the results

In conclusion, at 5.00 m ahead of both the electron and proton on their respective line of motion, there is no magnetic field created regardless of the particle. The magnitude of the magnetic field at this specific point is zero.