Question

A long straight wire carries a current of 1.5A an electron travels with a speed of 5×10^6cm/s parallel to the wire 10cm 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 created by the current in the wire is circular, centered around the wire. It is equal to$B = \frac{\mu_0 I}{2 \pi R}$, where $I$ is the current in the wire, $R$ is the distance from the wire and $\mu_0 = 4 \pi \cdot 10^{-7} \frac{T m}{A}$ is the permeability of the free space.

The force of the magnetic field on the electron is Lorentz force $\bold F= q \bold v \times \bold B$. Since electron is moving along the wire, $\bold v$ is perpendicular to $\bold B$, so $F = q v B \sin 90^{\circ} = q v B = \frac{\mu_o I q v}{2 \pi R} \approx 2.4 \cdot 10^{-20} N$.