An alpha particle charge 2e and an electron move in opposite

An alpha particle (charge +2e) and an electron move in opposite directions from the same point, each with the speed of 1.25 times 10^5 m/s (see the following figure). Find the magnitude and direction of the total magnetic field these charges produce at point P, which is 2.75 nm from each of them. Magnitude direction

Solution

the magnetic fields due to both paerticles are calculated as follows\":

   Balpha = {[u₀/4*pi][q*v*sin@]}/d^2

               = {[u₀/4*pi][2e*v*sin140]}/d^2

               = {[4*pi*10^-7 / 4*pi][2(1.6*10^-19)*(1.25x10^5)*sin140]}/(2.75x10^-9)^2

               =3.4e-4 T,   out of the page

Be = {[u₀/4*pi][q*v*sin@]}/d^2

               = {[u₀/4*pi][e*v*sin140]}/d^2

               = {[4*pi*10^-7 / 4*pi][(1.6*10^-19)*(1.25x10^5)*sin140]}/(2.75x10^-9)^2

               = 1.7e-4 T , out of the page

hence, the net magnetic field is,

   B = 3.4e-4 T + 1.7e-4 T = 5.1e-4 T = 0.51 e-3 T, out of the page