Speed of Light. Magnetic Field

Answers

1) $c = omega/k = 1/sqrt{(epsilon_0*mu_0)} =1/sqrt{(8.856*10^{-12}*4pi*10^{-7})}=299761783.5 m/s$

2) only the component of the speed perpendicular to B (magnetic field ) contributres to the magnetic force

$v = vi +vj =32*i+40*j m/s$

$F = -e(v times  B)$

x is the vector product between v and B

$F = e v j*B$   (in modulus)

this force is equal to the centripetal force

$F = m vj^2/R$

$e*vj*B = m*vj^2/R$

$R = mvj/(e*B) =9.1*10^{-31}*40000/1.6*10^{-19}*60*10^{-6} =3.792*10^{-3} m = 3.792 mm$

time T for one revolution is

$2*pi*R = vj*T$

$T = 2*pi*R/vj =2*pi*3.792*10^{-3}/40000=5.956*10^{-7} sec$

pitch of the path

$X = vi*T = 32000*5.956*10^{-7} =19.06*10^{-3} m =19.06 mm$

The electron is spiraling clockwise. See the picture below.