Asource injects an electron of speed v = 2.9 × 107 m/s into a uniform magnetic field of magnitude b = 1.7 × 10-3 t. the velocity of the electron makes an angle θ = 8.4° with the direction of the magnetic field. find the distance d from the point of injection at which the electron next crosses the field line that passes through the injection point.

r = 0.664 m.

Explanation:

Let's write the equation of the magnetic force, the blacks syndicate vectors

       F = q v x B

From this expression we see that the force is perpendicular to the velocity and the field, so it is a centripetal force, the modulus of the force is

      F = q v B sinT

We write Newton's second law

      F = m a

      a = v² / r

     q v B sinT = m v² / r

     r = m v / (q B sinT)

Let's calculate

     r = 9.1 10-31 2.9 107 / (1.6 10-19 1.7 10-3 sin8.4)

     r = 26.4 10-24 / 0.3973 10-22

     r = 0.664 m

This is the distance from where the electron penetrates

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j