Si un electrón en un haz de electrones experimenta una fuerza hacia abajo de 2.0×10'+,N mientras viaja en un campo magnético de 8.3 × 10'( T hacia el oeste, ¿cuál es la dirección y magnitud de la velocidad?

Magnitude = 1.51 ×10^6m/s

Direction is to the north

Explanation:

Question:

If an electron in an electron beam experiences a downward force of 2.0 × 10^-14N while traveling in a magnetic field of 8.3 × 10^-2T to the west, what is the direction and magnitude of the velocity?

Solution:

F= 2.0 × 10^-14N

Magnetic force = B = 8.3 × 10^-2T

Let the charge, q= 1.6×10^-19

F = Bqvsinθ

v = F/(Bqsinθ)

v = 2×10^-14/(8.3 × 10^-2 × 1.6×10^-19 ×sin90)

v = 2×10^-14/(13.28×10^-21 × 1)

v = 2/13.28 × 10^-14/10^-21

v = 0.1506 × 10^(-14+21)

v = 0.1506 × 10^7 = 0.151 × 10^7

v = 1.51 ×10^6m/s

Force is going down

magnetic field points west

Using the right and rule:

Point your fingers towards the west, and your thumb will point north. Since this is an electron, use the back of your hand( turn your hand backwards). The thumb points north.

Velocity direction is north

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied