Answer to Question #89982 in Electricity and Magnetism for Elias Gbagba

Question

An alpha particle (charge +2e) and an electron move in opposite directions from the same point, each with the speed of 2.50 X 105 m/s (Fig). Find the magnitude and direction of the total magnetic field these charges produce at point p. which is 1.75 nm from each of them.

Answer 1

Magnetic field of a moving charged particle:

$B=\frac{\mu_0}{4\pi}\frac{q[v,r]}{r^3}$

Suppose that an alpha particle flies to the right, and an electron to the left, then the field of the alpha particles and the electron will be directed in the “to us” direction and the total magnetic mole will be the sum.

$B_{tot}=\frac{\mu_0}{4\pi}\frac{q_{\alpha}[v_{\alpha},r]}{r^3}+\frac{\mu_0}{4\pi}\frac{q_{e}[v_{e},r]}{r^3}=\\ \frac{\mu_0}{4\pi}\frac{v}{r^2}(|q_{\alpha}|+|q_e|)=\\ \frac{4\pi\times10^{-7}}{4\pi}\frac{2.5\times10^5}{1.75\times10^{-9}}(3\times1.6\times10^{-19})= 6.86\times10^{-12}T$