Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 4.1.4

Interactive: Charged Particle in E and B

Home exercise 4.1.4

Charge in crossed fields

A particle of charge $q$ and mass $m$ enters with velocity $v = v a_{x}$ a region where a uniform electric field $E = - E a_{y}$ and a uniform magnetic field $B = - B a_{z}$ (into the page) are maintained. (a) Write the Lorentz force on the particle at the entrance and find the speed $v$ for which it moves on undeflected (velocity selector). (b) Now let $E = 0$ . Show that the motion is a circle, find its radius $r$ and the period $T$ of one revolution. (c) For a proton with $v = 1 0^{6}$ m/s and $B = 0.1$ T, evaluate $r$ and $T$ .

Try it yourself first.

Strategy

The Lorentz force with both fields present, evaluated at the entrance. For the circle, use the fact that a magnetic force is always perpendicular to $v$ , so it only turns the velocity, and Newton's law with the centripetal acceleration.

a) The velocity selector

At the entrance $v = v a_{x}$ and $B = - B a_{z}$ , so $v \times B = - v B (a_{x} \times a_{z}) = v B a_{y}$ , and

F = q (E + v \times B) = q (- E + v B) a_{y} .

The electric force pushes the particle down, the magnetic force up (for $q > 0$ , and both reverse together for $q < 0$ ). The particle is undeflected when they balance:

Answer

v = \frac{E}{B} independent of q and m

Faster particles are bent up, slower ones down: a velocity selector, which also provides ions of one speed to a mass spectrometer.

b) The circle

With $E = 0$ the force $F = q v \times B$ is always perpendicular to $v$ : it does no work, $∣ v ∣ = v$ stays constant, and a force of constant magnitude $q v B$ normal to the velocity produces uniform circular motion in the plane perpendicular to $B$ (Figure 4.1.28). Newton's law with the centripetal acceleration $v^{2} / r$ :

q v B = \frac{m v ^{2}}{r} ⟹ r = \frac{m v}{q B}, T = \frac{2 π r}{v} = \frac{2 π m}{q B}, f_{c} = \frac{1}{T} = \frac{q B}{2 π m} .

Answer

r = \frac{m v}{q B}, T = \frac{2 π m}{q B} (T does not depend on v or r)

c) A proton

r T = \frac{1.6726 \times 1 0 ^{- 27} \times 1 0 ^{6}}{1.602 \times 1 0 ^{- 19} \times 0.1} = 0.104 m = 10.4 cm, = \frac{2 π \times 1.6726 \times 1 0 ^{- 27}}{1.602 \times 1 0 ^{- 19} \times 0.1} = 6.56 \times 1 0^{- 7} s = 0.66 μ s,

i.e. $f_{c} = 1.52$ MHz per $0.1$ T ( $15.2$ MHz per tesla).

Home exercise 4.1.4(b). In a uniform field (into the page) the magnetic force keeps turning the velocity without changing its magnitude, and a positive charge runs counter-clockwise on a circle of radius mv/qB. The period does not depend on the speed: fast particles run on larger circles in the same time. — **Figure 4.1.28.** Home exercise 4.1.4(b). In a uniform field (into the page) the magnetic force keeps turning the velocity without changing its magnitude, and a positive charge runs counter-clockwise on a circle of radius $m v / q B$ . The period does not depend on the speed: fast particles run on larger circles in the same time.

At $v_{0} = E / B$ the proton passes undeflected. Try 'From rest' for the cycloid, or switch to Cyclotron to read $r$ and $T$ .

InteractiveLoading simulation…

Check

Units: $m v / (q B)$ is (kg m/s)/(C T) $=$ (kg m/s)/(C N s/(C m)) $=$ m. The independence of $T$ from the speed is what the cyclotron lives on: an alternating voltage of fixed frequency $f_{c}$ keeps kicking the ion as it spirals outward. At $v = 1 0^{6}$ m/s ( $0.3 %$ of $c$ ) the classical formulae are excellent. For electrons, or for protons above a few tens of MeV, $m$ must be replaced by $γ m$ . Note the direction: a positive charge in $B = - B a_{z}$ turns counter-clockwise seen from $+ z$ , since the force at the entrance is along $+ a_{y}$ (Figure 4.1.28).

Home exercise 4.1.4: Charge in crossed fields