# Deflecting a Beam of Electrons

1. Mar 26, 2016

1. The problem statement, all variables and given/known data

2. Relevant equations

W = -U for conservative forces
F = qV x B = qvBsinθ
W = Fd for conservative forces
K = 1/2mv^2
sinθ = 1 for 90° angles
E1 = E2 (conservation of mechanical energy)

3. The attempt at a solution

W = -U
-Fd = U

E1 = E2
K = U
K = -Fd = -qvBsinθd
K =-qvBd
-K/qd = Bv (1)

K = (1/2)mv^2
∴ v = sqrt(2K/m)
Substitute into (1)

-K/qd = B ⋅ sqrt(2K/m)
K^2/q^2d^2 = B^2 ⋅ (2K/m)
mK/2q^2d^2 = B^2
B = sqrt(mk/2q^2d^2)

Why is two in the denominator rather than in the numerator? Is my method invalid? If so, where can I improve? Thanks in advance.

2. Mar 26, 2016

### haruspex

The equation K=-Fd is for work done when a force advances distance d in tne direction of the force. Which way is the force here? Which way does the electron move?

3. Mar 26, 2016

The electron moves up and to the left in a parabolic trajectory. Therefore, there must be a force exerted up and a force exerted left which negates the electrons initial kinetic energy. How does this realization affect my derivation?

4. Mar 26, 2016

### haruspex

that shows the direction of the force changes over time. But the electron moves in a circular arc. Look at any part of that arc. What is the relationship there between the direction of the movement of the electron and the direction of its acceleration? What does that tell you about the direction of the force?

5. Mar 26, 2016

The force is variable, therefore my method is invalid. How should I approach this problem, then?

6. Mar 26, 2016

### Staff: Mentor

The motion follows a circular trajectory. What sort of force results in such a trajectory?

7. Mar 26, 2016

Centripetal

F = mv^2/r

8. Mar 27, 2016

Yes.