Question

A long straight wire carries a current of 1.5 A. an electron travels with a speed of 5 x 106 cm/s parallel to the wire 10 cm from it and in the same direction as the current. What force does the magnetic field of the current exert on the moving electron.

Answer 1

Magnetic field around the long straight wire carrying current is a well-known result, which can be obtained using Bio-Sawart law or Ampere's law: $B = \frac{\mu_0 I}{2 \pi R}$. The field is circular, and centered around the wire.

Magnetic field acts on the charged particle with Lorentz force: $\bold F = q \bold v \times \bold B$. Since, in our case the magnetic field is circular in the plane, perpendicular to the wire, and electron is moving along the wire, $\bold v$ is perpendicular to $\bold B$, so the absolute value of the cross product in the Lorentz force is just $v B$. Therefore, $F = q v B = \frac{q v \mu_0 I}{2 \pi R}$. Substituting given values for $I, v, R$ and permeability $\mu_0 = 4 \pi \cdot 10^{-7} \frac{T m}{A}$, obtain: $F = 2.4 \cdot 10^{-24} N$.