Question

An electron moves at 0.5 c (that is, 50% of the speed of light) in the negative y- direction (see 3D diagram z are all mutually perpendicular). There is a magnetic field of 2.5 mT in the z-direction. What is the force on the electron? (Give magnitude and direction!)

Answer 1

Given: Charge= -e , B(vector) = 2.5 mTin = 0.0025T and Velocity of charge(V) = (-0.5c) = -1.5*10⁸ m/s $\widehat{j}$ .

Given: change = - e, B (vector) = 2.5 m tins = 0 00 2.5 and velocity of change (v) = (-5c)^j jap = 1.5 times = - 10 ^8 m/s j^jap f is force vector, E (electrion change ) = (1 . 60 21 7662 x 10_-19 coulombs and velocity of light (C) = 3 times 10

$\underset{F}{\rightarrow}$ =q $(\underset{V}{\rightarrow}\times\underset{B}{\rightarrow})$ = $q*V*B\sin \theta$     (Here $\theta$ is 90⁰)

$\underset{F}{\rightarrow}$ =q $(\underset{V}{\rightarrow}\times\underset{B}{\rightarrow})$ = (-e)*(0.5c*0.0025)(-j $\times$ k) = 1.25 $\times$ 10^-3(e)*(c) $\widehat{i}$ = 2.002625 $\times$ 10^(-14) N in positive X-direction.

Here F is force vector, e(electronic charge) = (1.60217662 × 10^-19 coulombs) and velocity of light(c)= 3*10⁸ m/s.