Force on Moving Charges (Cyclotron & Motion)

PHYMagnetic Field & Forces

A charge moving in a magnetic field feels the Lorentz force $\vec{F}=q\vec{v}\times\vec{B}$ , which is always perpendicular to $\vec{v}$ and therefore does no work — it bends the path into a circle (or helix) without changing the speed. This gives the radius of circular motion and the speed-independent cyclotron frequency, the basis of the cyclotron accelerator.

ISC questions ask for the force on a moving charge, the radius of its path, the cyclotron frequency, and the maximum energy in a cyclotron.

Magnetic force on a moving charge

F = qvB\sin\theta

q

charge,

v

speed,

B

field,

\theta

angle between

\vec{v}

and

\vec{B}

; maximum when

\theta=90^{\circ}

, zero when

\theta=0^{\circ}

Radius of circular path

r = \dfrac{mv}{qB}

m

mass of particle,

v

speed perpendicular to

\vec{B}

; obtained from

qvB = mv^2/r

Cyclotron frequency and angular frequency

f = \dfrac{qB}{2\pi m}, \qquad \omega = \dfrac{qB}{m}

f

frequency in Hz,

\omega

angular frequency in

\mathrm{rad\,s^{-1}}

; both independent of speed and radius.

Radius after acceleration through p.d.

V

r = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}

Combines

qV = \tfrac12 mv^2

with

r=mv/qB

;

V

accelerating potential difference.

Maximum kinetic energy in a cyclotron

K_{max} = \dfrac{q^2 B^2 r_{max}^2}{2m}

r_{max}

maximum orbit (dee) radius; obtained from

K=\tfrac12 mv^2

with

v=qBr/m

The magnetic force is perpendicular to the velocity, so it does zero work: speed and kinetic energy are constant in a pure magnetic field.
A charge entering perpendicular to $\vec{B}$ moves in a circle; entering at an angle gives a helix (the parallel velocity component is unaffected).
Cyclotron frequency $f=qB/2\pi m$ is independent of speed and radius — the cornerstone that lets a fixed-frequency oscillator keep accelerating the particle.
Time period $T=2\pi m/qB$ is also speed-independent; the particle takes equal time per revolution regardless of how fast it goes.
For an electron $q=1.6\times10^{-19}\ \mathrm{C}$ and $m=9.1\times10^{-31}\ \mathrm{kg}$ ; for a proton $m=1.67\times10^{-27}\ \mathrm{kg}$ .
When a charge is accelerated through a p.d. $V$ , its kinetic energy is $qV$ , giving $v=\sqrt{2qV/m}$ before it enters the field.
Maximum energy in a cyclotron is limited by the dee radius $r_{max}$ and field $B$ ; the relativistic mass increase eventually breaks the resonance condition.

Where the marks go

Forgetting the $\sin\theta$ factor in $F=qvB\sin\theta$ , e.g. using full $qvB$ when the charge moves at $30^{\circ}$ .
Thinking the cyclotron frequency depends on speed, radius or kinetic energy — it depends only on $q/m$ and $B$ .
Believing the magnetic force changes the particle's speed or energy; it only changes the direction.
When the charge is accelerated through $V$ first, forgetting to compute $v$ from $qV=\tfrac12 mv^2$ before applying $r=mv/qB$ .

How the board asks it

Numericalradius of circular path and cyclotron frequency
A proton moving with a speed of $4\times10^{6}\ \mathrm{m\,s^{-1}}$ enters a uniform magnetic field of $0.5\ \mathrm{T}$ at right angles to the field. Calculate the radius of its circular path and the frequency of its revolution. (Mass of proton $=1.67\times10^{-27}\ \mathrm{kg}$ )
Numericalmaximum kinetic energy from dee radius and field
In a cyclotron of dee radius $0.5\ \mathrm{m}$ and magnetic field $1.2\ \mathrm{T}$ , calculate the maximum kinetic energy (in $\mathrm{MeV}$ ) to which a proton can be accelerated. (Mass of proton $=1.67\times10^{-27}\ \mathrm{kg}$ )
Derive / proveradius of circular path and speed-independent period
Obtain an expression for the radius of the circular path of a charged particle moving perpendicular to a uniform magnetic field, and hence show that its period of revolution is independent of its speed.
Give reasonsmagnetic force does zero work
A charged particle moving in a uniform magnetic field follows a circular path with constant speed. Give reasons why the magnetic force does no work on the particle and yet changes its direction of motion.
Applicationforce on a moving charge $F=qvB\sin\theta$
An electron enters a uniform magnetic field $\vec{B}$ with its velocity making an angle of $30^{\circ}$ with the field. Describe the nature of its subsequent path and state the expression for the magnetic force acting on it.
Define / statecyclotron frequency $f=qB/2\pi m$
State what is meant by cyclotron frequency and explain why it is independent of the speed of the charged particle.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.