an electron

[logistic_guy]

An electron is moving in a plane that also contains a long straight current-carrying wire. The electron is heading at a $\displaystyle 45^{\circ}$ angle toward the wire, with a speed of $\displaystyle 3.4 \times 10^6 \ \text{m/s}$, when it is $\displaystyle 50 \ \text{cm}$ away. The electron reaches only as close as $\displaystyle 1 \ \text{cm}$, before being repelled away, always moving in the same plane. What is the current in the wire?

 

[Gadsilla]

Explore Exciting Connections for No-Strings Attached Encounters in Your City

 

[khansaheb]

An electron is moving in a plane that also contains a long straight current-carrying wire. The electron is heading at a $\displaystyle 45^{\circ}$ angle toward the wire, with a speed of $\displaystyle 3.4 \times 10^6 \ \text{m/s}$, when it is $\displaystyle 50 \ \text{cm}$ away. The electron reaches only as close as $\displaystyle 1 \ \text{cm}$, before being repelled away, always moving in the same plane. What is the current in the wire?

show us your effort/s to solve this problem.

 

[logistic_guy]

show us your effort/s to solve this problem.

This is a very complicated problem. Rigorous understanding of Magnetism is necessary. Besides that we have to understand the geometry of the problem. If we have a long straight current-carrying wire, we know that it produces a magnetic field near it equal to:

$\displaystyle B = \frac{\mu_0 I}{2\pi R} = \frac{\mu_0 I}{2\pi y}$

If a particle is moving in this magnetic field, a magnetic force will act on it equal to:

$\displaystyle F = qvB = evB$

Since the electron is moving in the same plane of the wire, the magnetic field must be in a plane perpendicular to this plane which means that the velocity and the force are perpendicular to each other in the same plane of the wire. Because of this geometry, the magnitude of the velocity will not change, it is only that its components will change.



From elementary physics, we know that the force is equal to:

$\displaystyle F = ma$

This force has two components.

$\displaystyle F_x = ma_x = m\frac{d v_x}{dt}$

$\displaystyle F_y = ma_y = m\frac{d v_y}{dt}$

We don't have enough information in the $\displaystyle x$ direction, so we will work in the $\displaystyle y$ direction. As the electron gets closer to the wire, its velocity in the $\displaystyle y$ direction decreases and its velocity in the $\displaystyle x$ direction increases. Therefore, at $\displaystyle y = 1 \ \text{cm}, v_y = 0 \ \text{and} \ v = v_x = 3.4 \times 10^6 \ \text{m/s}$. In fact $\displaystyle |v| = 3.4 \times 10^6 \ \text{m/s}$ at any point.

This information is very important and it is the key to solve this problem. Let us work on the force in the $\displaystyle y$ direction.

$\displaystyle F_y = m\frac{d v_y}{dt}$

$\displaystyle F\cos \theta = m\frac{d v_y}{dt}$

$\displaystyle evB\cos \theta = m\frac{d v_y}{dt}$

We know that $\displaystyle v_y = v\sin \theta$ then $\displaystyle d v_y = v \cos \theta \ d\theta$ and from the chain rule, we get:

$\displaystyle \frac{d v_y}{dt} = \frac{d v_y}{dy}\frac{dy}{dt} = \frac{v\cos\theta \ d\theta}{dy}v_y = v^2 \sin\theta \cos\theta\frac{d\theta}{dy}$

$\displaystyle evB\cos \theta = m\frac{d v_y}{dt}$

$\displaystyle ev\left(\frac{\mu I}{2\pi y}\right)\cos \theta = mv^2 \sin\theta \cos\theta\frac{d\theta}{dy}$


$\displaystyle \int_{0.5}^{0.01}e\left(\frac{\mu I}{2\pi y}\right) \ dy \ = \int_{\frac{\pi}{4}}^{0}mv \sin\theta \ d\theta$


$\displaystyle e\left(\frac{\mu I}{2\pi}\right)\ln y\bigg|_{0.5}^{0.01} \ = -mv \cos\theta \bigg|_{\frac{\pi}{4}}^{0}$


$\displaystyle e\left(\frac{\mu I}{2\pi}\right)(\ln 0.5 - \ln 0.01) \ = mv \left(\cos 0 - \cos\frac{\pi}{4}\right)$


$\displaystyle I = \frac{2\pi mv\left(\cos 0 - \cos\frac{\pi}{4}\right)}{e\mu (\ln 0.5 - \ln 0.01)} = \frac{2\pi (9.11 \times 10^{-31})(3.4 \times 10^{6})\left(\cos 0 - \cos\frac{\pi}{4}\right)}{(1.6 \times 10^{-19})(4\pi \times 10^{-7}) (\ln 0.5 - \ln 0.01)} \approx 7.2 \ \text{A}$