Source: High school physics (Chinese)
Problem Sets:
Problem
A long straight wire carries a current of $I = 20.0$ A. An electron moves with speed $v = 5.0\times10^6 \text{ m/s}$ at a distance $r = 1.0 \text{ cm}$ from the wire. Find the magnitude and direction of the magnetic force on the electron when it moves in each of the following directions:
- Radially away from the wire.
- Parallel to the wire and in the same direction as the current.
- Perpendicular to the wire and tangent to the circular field line at the electron's location.
(1) $F = 3.2\times10^{-16}$ N, directed opposite to the wire's current.
(2) $F = 3.2\times10^{-16}$ N, directed radially away from the wire.
(3) $F = 0$.
At the electron's location, the wire's magnetic field has magnitude
$$B = \frac{\mu_0 I}{2\pi r} = \frac{(2\times10^{-7})(20.0)}{0.010} = 4.0\times10^{-4}\text{ T},$$with $\vec{B}$ tangent to the circle around the wire (perpendicular to both the wire and the radial direction).
Lorentz force on the electron: $\vec{F} = -e\,\vec{v}\times\vec{B}$. Whenever $\vec{v}\perp\vec{B}$,
$$F = evB = (1.6\times10^{-19})(5.0\times10^6)(4.0\times10^{-4}) = 3.2\times10^{-16}\text{ N}.$$(1) $\vec{v}$ radially outward $\perp \vec{B}$ (tangential). Evaluating $-e\,\vec{v}\times\vec{B}$: $\vec{F}$ points opposite to the wire's current direction, $F = 3.2\times10^{-16}$ N.
(2) $\vec{v}$ along current $\perp \vec{B}$. Evaluating $-e\,\vec{v}\times\vec{B}$: $\vec{F}$ points radially away from the wire, $F = 3.2\times10^{-16}$ N.
(3) $\vec{v}$ tangent to the field circle means $\vec{v}\parallel\vec{B}$, so $\vec{v}\times\vec{B}=0$ and $\vec{F}=0$.