Motion of Charged Particles in Electromagnetic Fields

Abstract

One of the most important applications of the electric and magnetic fields deals with the motion of charged particles. For instance, in experimental nuclear fusion reactors the study of the plasma requires the analysis of the motion, radiation, and interaction, among others, of the particles that forms the system. In biomedicine the use of accelerators, like the cyclotron, allows the preparations of compounds to be employed in diagnostics, such as the FDG (see Problem 11.5), which is used as a tracer in the detection of some kinds of cancer and body diseases by means of the PET technique. Or in a closer case, the Earth, where its magnetic field, acting like a particle mirror, protects us against many of the cosmic rays.

Appendices

Solved Problems

Problems A

11.1. :

A proton of mass $m=1.672 \times 10^{-27}$ kg and electrical charge $q\,{=}\,e \,{=}\,1.602 \times 10^{-19}$ C is left without an initial velocity in a homogeneous electric field $E=20\,\mathrm{V/m}$. The velocity that the proton acquires and the distance travelled when the elapsed time is 0.08 s are required.

Solution

As the charge and the electric field are known, the force that is exerted on the particle can be calculated. With the force calculated and the mass known, the fundamental equation of dynamics allows the acceleration to be calculated. From the value of the acceleration it is possible to obtain the velocity and the displacement.

The equations (11.7)–(11.9) can be applied directly, in which, the initial velocity is $v_{0x}= v_{0y}= v_{0z}=0$. If the origin of coordinates is taken as the starting point of the proton, and the axis OZ coincides with the direction of the electric field, then:

$$ \begin{array}{l} v_x =0\quad \Rightarrow \quad x=0, \\ v_y =0\quad \Rightarrow \quad y=0, \\ \end{array} $$

$$ \begin{array}{l} \displaystyle {v_z =\frac{1.602\times 10^{-19} \times 20 \times 0.08}{1.672 \times 10^{-27}}=1.5330 \times 10^{8}\,\text {m/s}}, \\ \displaystyle {z=\frac{qE}{2m}t^{2}=\frac{1.602 \times 10^{-19} \times 20}{2 \times 1.672 \times 10^{-27}} 0.08^{2} \text {m} = 6.132 \times 10^{6}\,\text {m}}\,. \end{array} $$

As the obtained velocity is about half the speed of light, these results can only be taken as approximate.

11.2. :

At a point in space there is an electric field E in the direction of the coordinate axis OX, a magnetic field B in the direction of the axis OY, and a particle of charge q moving with velocity v in the direction of the axis OZ (Fig. 11.10). Calculate the components of the force along the three coordinate axes.

Solution

The force acting on the particle is the Lorentz force

$$ \mathbf{F}=q\mathbf{E}+q\mathbf{v}\times \mathbf{B}\,. $$

Since E has the direction of OX axis, then the components of q E are, respectively: qE, 0, and 0.

The vectorial product v $\times$ B has the opposite direction to OX axis, and its respective components are: - qvB, 0, and 0.

Therefore, the components of the force are, respectively: (qE–qvB), 0, and 0 and can be written

$$ \mathbf{F}=q\left( {E-vB} \right) \mathbf{u}_x . $$

11.3. :

An ion of charge $q=e=1.602 \times 10^{-19}$ C and mass $m=1.50 \times 10^{-25}$ kg is impelled with a velocity $v=$100000 m/s perpendicular to a homogeneous electric field $E=$ 3 V/m. Calculate the velocity acquired by the ion during the first 0.4 s and draw the trajectory.

Fig. 11.10

Particle with velocity v

Solution

The force on the particle can be calculated from data q and E. The acceleration is calculated from the force and the mass, and hence the velocity and the trajectory.

Drawing coordinate axes with their origin at the initial position of the particle, with the OZ axis in the direction of the electric field and with the OX axis in the direction of the initial velocity, as shown in Fig. 11.11. With respect to these axes, the data can be written thus:

$$ v_{0x}={v}=100000\,\mathrm{m/s}, v_{0y}=0, v_{0z}=0. $$

The distance travelled along axis OX in 0.4 s is, according to (11.7),

$$ x_{0.4} =v_{0x} t=100000\times 0.4\;\text {m}=40000\;\text {m}. $$

The velocity along axis OZ is, according to (11.9),

$$ v_z =\frac{qE}{m}t+v_{0z} =\frac{qE}{m}t=\frac{1.602 \times 10^{-19}\times 3}{1.50 \times 10^{-25}}0.03\;\text {m/s}\,=\,9.612\times {10}^{4}\;\text {m/s}. $$

Directly applying (11.10) gives the trajectory

$$ z=\frac{qE}{2mv^{2}}x^{2}=\frac{1.602 \times 10^{-19}\times 3}{2 \times 1.50 \times 10^{-25}\times 100000^{2}}x^{2}=1.602\times 10^{-4}x^{2}. $$

The graphical representation is given as a continuous line in Fig. 11.11.

11.4. :

An electric field has the direction of axis OY and its modulus varies with the point of space in the form $E=E_{0}+kz$, where k is a constant. At the initial instant, a particle of charge q and mass m is impelled from the origin of the coordinates, with a velocity $\mathbf{v}_{0}$ parallel to axis OZ (Fig. 11.12). Find the distance of the particle to the origin of the coordinates at instant t.

Fig. 11.11

Reference of frame

Solution

Since the electric field, the charge, and the mass are known, it is possible to calculate the acceleration, the velocity and the components of the displacement along the axes. Given this displacement, the distance can be determined.

The field has the components $E_{x}=0$, $E_{y}=E_{0}+kz$, and $E_{z}=0$, which cause the respective accelerations $a_{x}=0$, $a_{y}=q(E_{0}+kz)/m$, and $a_{z}=0$. Therefore:

$$ \begin{array}{l} \displaystyle {\frac{dv_x }{dt}=0\quad \Rightarrow \quad v_x =C_1 =0\quad \Leftrightarrow \quad \frac{dx}{dt}=0\quad \Rightarrow \quad x=C_2 =0}, \\ \displaystyle { \frac{dv_y }{dt}=\frac{q}{m}\left( {E_0 +kz} \right) }, \\ \displaystyle { \frac{dv_z }{dt}=0\quad \Rightarrow \quad v_z =C_3 =v_0 \quad \Leftrightarrow \quad \frac{dz}{dt}=v_0 \quad \Rightarrow \quad z=v_0 t+C_3 =v_0 t}. \\ \end{array} $$

By substituting the third result into the second, this is transformed into

$$ \begin{array}{l} \displaystyle { \frac{dv_y }{dt}=\frac{q}{m}\left( {E_0 +kv_0 t} \right) \quad \Rightarrow \quad v_y =\frac{q}{m}\left( {E_0 t+\frac{kv_0 }{2}t^{2}} \right) +C_4 =\frac{q}{m}\left( {E_0 t+\frac{kv_0 }{2}t^{2}} \right) } \\ \displaystyle { \Rightarrow y=\frac{q}{m}\left( {\frac{E_0 t^{2}}{2}+\frac{kv_0 }{2\times 3}t^{3}} \right) +C_5 =\frac{q}{2m}\left( {E_0 t^{2}+\frac{kv_0 }{3}t^{3}} \right) }. \\ \end{array} $$

The distance d to the centre is therefore

$$ d=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{\frac{q^{2}}{4m^{2}}\left( {E_0 t^{2}+\frac{kv_0 }{3}t^{3}} \right) ^{2}+\left( {v_0 t} \right) ^{2}}.\quad $$

11.5. :

Electrons, $q=-e$, are used as a test charge to determine a field B. This field can be considered homogeneous, stationary, perpendicular to the plane of Fig. 11.13, and confined to the hatched area. The electrons are accelerated starting from rest when passing through plates between which there is a difference of potential V’=$V_{2}-V_{1}$.

(a) Determine the modulus and direction of B if, after a certain route through the interior of the magnetic field, the point of impact of the electrons on the screen (plane $x=0$) is at (0, a, 0). The mass and the charge of the electron are assumed to be known.

(b) Later, at a point P of the previous field B, two charged particles are impelled with the same velocity perpendicular to the field (Fig. 11.14). These particles have equal mass. The figure shows the trajectory followed by each particle, recorded by means of a Wilson cloud chamber (basically it contains gas and water vapour). Give reasons for the difference between these particles and explain what may cause the progressive reduction of the radius of curvature.

Fig. 11.12

Electric field

Fig. 11.13

Experimental set-up with magnetic field

Fig. 11.14

Trajectory of the two particles

Solution

(a) The principle of conservation of energy establishes that the sum of the kinetic and potential energies in 1 and 2 is the same, that is

$$ 0+qV_1 =\frac{1}{2}mv^{2}+qV_2 \quad \Rightarrow \quad -eV_1 =1/2mv^{2}-eV_2 \quad \Rightarrow \quad v=\sqrt{\frac{2eV'}{m}}\,. $$

If $q>0$, then the impact is to the right, and therefore the half circumference travelled within the magnetic field has its centre at (0, a/2, 0). Since the centripetal acceleration, caused by B, points towards the centre of the circumference, then so does the force $q \mathbf{v} \times \mathbf{B}= \mathbf{-ev} \times \mathbf{B}$, and therefore $\mathbf{v} \times \mathbf{B}$ points away from the centre. Hence B points out of the paper. Put another way, at the moment when the electron enters the field,

$$\begin{aligned} \mathbf{F}=-e\mathbf{v}\times \mathbf{B}\quad&\Rightarrow \quad F_y \mathbf{u}_y =-ev_x \mathbf{u}_x \times B_z \mathbf{u}_z =ev_x B_z \mathbf{u}_y \quad \Rightarrow \quad F_y =ev_x B_z \\&\Rightarrow \quad B_z =\frac{F_y }{ev_x }>0. \end{aligned}$$

Moreover, the fundamental equation of dynamics gives

$$ \frac{mv^{2}}{R}=ev_x B_z =evB\quad \Rightarrow \quad B=\frac{mv}{eR}=\frac{m}{ea/2}\sqrt{\frac{2eV'}{m}}=\frac{2}{a}\sqrt{\frac{2mV'}{e}}\,. $$

(b) The main difference is that the initial centres of curvature are one to each side, therefore the acceleration is equal except for the direction, and hence the force is equal except for the direction, and therefore the charges must be of opposite signs. The particle to the left must have a negative charge and the particle to the right a positive charge . This allowed experimentally discovery of a new particle: the positron ${ }_{+1}^{\,\,0} e$. The reduction of the radius of curvature is due, with high probability, to the reduction of velocity, caused by the collision of the particles with the gas molecules in the cloud chamber .

The existence of the positron was predicted by P. Dirac in 1931, and discovered experimentally by C. Anderson in 1932 while studying cloud chamber photographs of cosmic rays . The positron has the same mass and magnitude of charge, but opposite in sign, as the electron. Rigorously speaking, it constitutes the antiparticle of the electron and it is antimatter. The behaviour of the antimatter is not as usual as thought. In fact, when an electron (matter) coincides with a positron (antimatter ) in a region of the space they annihilate each other. As a result two photons (gamma rays) appear moving in opposite directions with energies of 511 keV (momentum conservation law). At first sight it may be thought that the study of the antimatter is only important for the physicists that investigate in quantum field theory, but that is completely wrong. The study of the elementary particles performs a fundamental role in great variety of subjects, such as biology , medicine, chemistry , and of course Physics. By way of illustration, suffice it to say, that the application of the positrons for the diagnostic of some diseases has been used since 1969 (the first time in USA) by means of the PET technique (Positron Emission Tomography ). More specifically, by this procedure it is possible to diagnose cancer, degenerative anomalies such as Alzheimer and Parkinson , metabolic disorders, and epilepsy, among others.

PET is a non-invasive method which employs chemical compounds labelled with radioisotopes of short haf-life time, like $^{11}$C, $^{13}$N, $^{15}$O and $^{18}$F. These compounds are called tracers and are injected into the body in order to measure where its activity is greatest. The election of the tracer depends on the disease to be investigated. However, one of the most employed tracer is FDG (Fluorodeoxyglucose). As the tumours consume more energy than normal cells, the FDG accumulates more in the regions where the body needs more energy. Due to the fact that the $^{18}$F is introduced in the molecule it disintegrates (beta plus decay ($\upbeta ^{+}), { }_1^1 p\rightarrow { }_0^1 n+{ }_{+1}^{\,\,0} e+{ }_0^0 \nu )$ emitting one positron (${ }_{+1}^0 e)$ which annihilates with one of the electrons of the surrounding matter leading to two photons. By using photomultiplier-scintillator detectors located on opposite sides, and computerized tomographic reconstruction based on correlation direction and time coincidence of the photons emitted, it is possible to obtain an image of the regions of more activity. From a medical viewpoint this technique is very sensitive for detecting the activity zones but not the morphology of the tumours. For this reason the new PET-machines bring an incorporated CT $^{1}$ ¹ scanner, which allows a good reconstruction of the region. As a result by means of the fusion of both data, a precise location of the tumour and its possible malignity are obtained.

11.6. :

A conductive strip (Hall probe) is located in a region of space where there is a known magnetic field B, and voltage $V_{H}$ is measured when the plane of the tape is perpendicular to the magnetic field. The probe is turned an angle $\theta$ around the axis of symmetry parallel to the longest edge. Calculate the Hall voltage that will be measured after the turn.

Solution

Fig. 11.15

Specimen

In the first position of the probe, the voltage is given by (11.17)

$$ V_H =wvB\quad \Rightarrow \quad wv=V_H /B\,. $$

In the second position (see Fig. 11.15), in permanent regime, the carrying charges, if positive, move in the direction of the vector density of current j and are subjected to the magnetic field and the electric field caused by the charges that have been deposited at the edges. The resultant of the forces throughout the width of the tape must be null. Therefore

$$ qE=qvB\cos \theta \quad \Rightarrow \quad E=vB\cos \theta . $$

The resultant voltage is obtained from

$$ E=\frac{V_{H}{}^{\prime }}{w}\quad \Rightarrow \quad V_{H}{}^{\prime } =wE=wvB\cos \theta \quad \Rightarrow \quad V_{H}{}^{\prime }=\frac{V_H }{B}B\cos \theta =V_H \cos \theta . $$

11.7. :

In a region of space, an electric field E and a magnetic field B are parallel and homogeneous. A particle with charge q and mass m is impelled with velocity $\mathbf{v}_{0}$ perpendicular to the fields. Calculate the advance made in the first turn and in the second turn.

Solution

If the OZ axis is drawn in the common direction of the fields and the origin of the coordinates is located at the point of release of the particle, then Fig. 11.16 shows the results. With the data given, the acceleration, velocity and displacement can be calculated. As demonstrated in the theoretical introduction (11.20), the electric field causes an acceleration along the axis OZ of value

$$ a_z =\frac{qE}{m}\quad \Rightarrow \quad v_z =\frac{qE}{m}t\quad \Rightarrow \quad z=\frac{qE}{2m}t^{2}\,. $$

Moreover, the rotation period caused by the magnetic field, (11.15), is

$$ T=\frac{2\pi m}{qB}\,. $$

Substituting this value of time into the previous equation gives the distance travelled in the direction of OZ axis in the first turn thus

$$ L_1 =\frac{qE}{2m}T^{2}=\frac{qE}{2m}\frac{2^{2}\pi ^{2}m^{2}}{q^{2}B^{2}}=\frac{2\pi ^{2}mE}{qB^{2}}\,. $$

In the time spent in the two turns, 2T, the distance travelled in the common direction is

$$ L_2 =\frac{qE}{2m}\left( {2T} \right) ^{2}=\frac{qE}{2m}\frac{2^{2}\times 2^{2}\pi ^{2}m^{2}}{q^{2}B^{2}}=4\frac{2\pi ^{2}mE}{qB^{2}}\,. $$

Note how the distance $L_{2}$ is not double but four times the distance $L_{1}$ in only double the time. The trajectory is not a simple helix but a kind of helix whose pitch increases with time.

11.8. :

Two isotopes of electrical charge $q=e=1.602 \times 10^{-19}$ C and masses $m_{1}=1.673 \times 10^{-26}\,\mathrm{kg}$ and $m_{2}=1.743\times 10^{-26}\,\mathrm{kg}$, respectively, enter the mass spectrometer described in Sect. 11.6. The electric field applied in the velocity selector is $E=1000$ V/m and the magnetic fields are equal in the whole device, $B=0.02$ T. Calculate the velocity of the isotopes on their arrival at the detector and the point where they can be detected.

Solution

Fig. 11.16

Electric and magnetic field

By observing the figure of Sect. 11.6, and applying (11.22), the velocity of the isotopes that cross the exit orifice of the velocity selector is calculated

$$ v=E/B_1 =1000/0.02\;\text {m/s}=50000\;\text {m/s}. $$

Since there is only one magnetic field inside the mass selector, and the modulus of velocity does not vary, then the arrival at the detector is with the velocity of 50000 m/s.

The radius of curvature is obtained by applying (11.23):

$$ R=\frac{mE}{qB_1 B_2 }=\frac{mE}{qB^{2}}\,. $$

Therefore, for the isotopes of mass $m_{1}$ and $m_{2,}$ the respective radii are: 

$$ R_1 =\frac{m_1 E}{qB^{2}}=\frac{1.673\times 10^{-26}\times 1000}{1.602\times 10^{-19}\text {C}\times 0.02^{2}}=0.2611\;\text {m}, $$

and

$$ R_2 =\frac{m_2 E}{qB^{2}}=\frac{1.743\times 10^{-26}\times 1000}{1.602\times 10^{-19}\text {C}\times 0.02^{2}}=0.2720\;\text {m}. $$

The points where they can be detected are at the respective distances $2R_{1}$ and $2R_{2}$.

11.9. :

A cyclotron of radius R has a space L between its D’s, such that $L\ll R$. There is a magnetic field B perpendicular to the plane of the cyclotron . A difference of potential $V=V_{0}\mathrm{cos} (\omega _{i} t)$ is applied between the D’s, where $\omega _{i}$ is the suitable angular frequency value for each particle. Two different types of particles, of identical positive charges but of respective masses $m_{1}$ and $m_{2}$, are impelled sequentially. (a) In a first experiment, the particle of mass $m_{1}$ and negligible initial velocity is accelerated. In a second experiment, the particle with mass $m_{2}$ is accelerated. Calculate the revolutions given by each particle. (b) Determine, by reasoning, the amount of energy supplied by the magnetic field to each of the particles. (c) Obtain the period of rotation of the particle of mass $m_{1}$ when the semicircular trajectory of radius R/2 is described and compare it with the period corresponding to the trajectory of the last cycle where the radius is R.

Solution

(a) The fundamental equation of dynamics for the particle travelling the final semi-circumference and projected on a radius and towards the centre:

$$ qv_f B=\frac{mv_{f}{}^{2}}{R}\quad \Rightarrow \quad v_f =\frac{qBR}{m}\,. $$

Energy that the electric field between the D’s contributes for each cycle (double pass): $2 qV_{0}$.

Applying the principle of conservation of energy to Nrevolutions, where they reach the final velocity $v_{f}$, we have

$$ \frac{1}{2}mv_{f}{}^{2}-0=N2qV_{0} \quad \Rightarrow \quad N=\frac{m}{4qV_{0} }v_f ^{2}=\frac{m}{4qV_{0} }\left( {\frac{qBR}{m}} \right) ^{2}=\frac{qB^{2}R^{2}}{4mV_{0} }\,. $$

For each particle, substitute m for $m_{1}$ or $m_{2}$ accordingly. Since the mass is in the denominator, the greater the mass, the fewer the number of cycles traced by the particle.

(b) The energy contributed by the magnetic field can be calculated by means of the work carried out by the force that the magnetic field exerts on the particle, which is

$$ W=\int _1^2 {\mathbf{F}.d\mathbf{r}=} \int _1^2 {q\mathbf{v}\times \mathbf{B}.d\mathbf{r}=} 0, $$

since v $\times$ B is perpendicular to v, that is, to the trajectory, and d r is tangent to the trajectory.

(c) The period is calculated by applying (11.15)

$$ T=\frac{2\pi m}{qB}, $$

and therefore the period depends on the mass of the particle, but is independent of the radius of the semi-circumference that it travels.

11.10. :

An electron with null velocity is injected into a betatron at distance $R=0.2$ m from its centre. The magnetic field varies from $B=0$ to $B= B_{max}=$ 0.005 T. Calculate the final energy of the electron.

Solution

The velocity at any instant, and the final velocity reached by the electron are obtained from (11.14):

$$ v=\frac{qRB}{m}\quad \Rightarrow \quad v_{\max } =\frac{qRB_{\max } }{m}\,. $$

Substituting the data from the statement and from the table of constants yields

$$ v_{\max } =\frac{1.602\times 10^{-19}\times 0.2\times 0.005}{9.109\times 10^{-31}}\frac{\text {m}}{\text {s}}=1.759\times 10^{8}\frac{\text {m}}{\text {s}}\,. $$

Since this velocity is close to the speed of light in a vacuum, it is not very reliable.

The kinetic energy acquired is estimated by means of substitution of this maximum value into the expression of the kinetic energy:

$$ E_k =\frac{1}{2}mv_{\max }^{^{2}} =\frac{1}{2}9.109\times 10^{-31}\times \left( {1.759\times 10^{8}} \right) ^{2}\text {J}=1.409\times 10^{-14}\text {J}. $$

11.11. :

An electron is pulled by the photoelectric effect , with negligible velocity, from the inner face of the negative plate of a flat capacitor. The separation between the plates is $D=2$ cm and the difference of potential between them is such that it is at the point of producing a disruptive discharge. Calculate: (a) the energy of an electron on being pulled; (b) the velocity acquired by an electron before colliding against the positive armature supposing that there are no collisions against air molecules. Take the data for m and q from the table of physical constants. The dielectric strength of the air is $E=30000$ V/cm.

Solution

(a) The potential energy is calculated by the work of the force that the electric field applies to the charge:

$$ E_p =\int _1^2 {q\mathbf{E}.d\mathbf{r}} =\int _1^2 {qE.dr} =-e(V_1 -V_2 )\equiv eV. $$

The maximum allowable difference of potential is

$$ V=ED. $$

Therefore the energy that it has is

$$ E_{p} = eED \, E_p =eED=1{.}602 \times 10^{-19}\times 30000\times 0.02\, \text {J}=961\times 10^{-19} \text {J}. $$

(b) According to the principle of conservation of energy we can write

$$\begin{aligned} eED=\frac{1}{2}mv^{2}\quad \Rightarrow \quad v&=\sqrt{\frac{2eED}{m}}=\sqrt{\frac{2\times 1{.}602\times 10^{-19}\times 30000\times 0.02}{9.107\times 10^{-31}}}\text {m/s}\\&=1.453\times 10^{7}\text {m/s}, \end{aligned}$$

which is, approximately, 5 % of the speed of light; therefore, the classical solution of the problem can give an approximated result, although not exact.

11.12. :

A synchrotron is formed by an annular vacuum tube of mean radius R (Fig. 11.17). Electrons are required to be accelerated to high velocities while maintaining the radius of the orbit. There is a magnetic field inside the ring, perpendicular to its plane. (a) Given the values for the energy $E_{r}$ and field B at a certain instant, calculate the period T. (b) An accelerating alternating voltage V of constant period Tis applied. Calculate the increase of B to compensate an increase of energy $\Delta E_{r}$ in a cycle. Given that it is desired to cause an increase per unit time of value $\Delta E_{r}/\varDelta t$, calculate the rapidity of the increase of B with time that is needed.

Solution

Fig. 11.17

The synchrotron

(a) Since the velocity to be attained is high, it is necessary to apply the formulae of relativistic mechanics. The period is obtained from (11.53)

$$ T=\frac{2\pi }{\omega }=\frac{2\pi E_r }{\text {e}c^{2}B}\,. $$

(b) From this expression B is obtained in terms of $E_{r}$

$$ B=\frac{2\pi }{\text {e}c^{2}T}E_r\,. $$

Therefore, in this problem, B only depends on the variable $E_{r}$. The $E_{r}$ is increased by $\varDelta E_{r}$ in each pass by the accelerating electric field due to V, and hence the magnetic field must be increased by

$$ \Delta B=\frac{2\pi }{\text {e}c^{2}T}\Delta E_r, $$

and the increase of B per unit of time is

$$ \frac{\Delta B}{\Delta t}=\frac{\Delta B}{T}=\frac{2\pi }{\text {e}c^{2}T^{2}}\Delta E_r . $$

Problems B

11.13. :

A beam of protons , of charge e, homogeneously distributed within a very long cylinder with n protons per unit volume, moves with velocity v along the cylinder axis. Calculate: (a) the electric field existing at distance r from the axis; (b). the magnetic field at this point; (c) the outward radial component of the resultant of the forces on one of the protons.

Solution

(a) As the distribution of charges is known, Gauss’s theorem can be applied to any closed surface.

Since there is an axis of symmetry, it is advantageous to consider a cylindrical surface of radius r, concentric with the charge conducting cylinder limited by bases distanced from each other in L (Fig. 11.18).

The flow of the electric field through the specified cylinder is

$$ \int {\mathbf{E }.d\mathbf{S }} =\int \limits _{lat} {E.dS} \cos 0^{\circ }+\int \limits _{ends} {E.dS} \cos 90^{\circ }=E\int \limits _{lat} {dS} =E2\pi rL, $$

where it has been taken into account: (1) the distributive property of the integral, (2) that the electric field is of radial direction due to the symmetry and that E is the component of E in the outward radial direction.

The electrical charge density is $\rho \equiv dq/d{Vol}$, therefore the charge within the cylinder is

$$ \int {\rho dVol=} \rho \int {dVol=\rho \pi r^{2}L} =ne\pi r^{2}L. $$

Gauss’s theorem establishes that

$$ E2\pi rL=\frac{ne\pi r^{2}L}{\varepsilon _0 }\quad \Rightarrow \quad E=\frac{ner}{2\varepsilon _0 }\,. $$

(b) As the system of currents is known and there is cylindrical symmetry, Ampère’s law may be applied to a circumference of radius r concentric with the axis of the cylinder.

The circulation of field B along the circumference is

$$ \oint {\mathbf{B}.d\mathbf{l}=} \oint {B_\phi dl=} B_\phi \oint {dl=} B_\phi 2\pi r=B2\pi r, $$

therefore, through symmetry, the only non-null component of B is the tangent to the circumference and B is its projection on d l.

Fig. 11.18

Cylindrical region with protons

The current density is

$$ \mathbf{j}=\rho \mathbf{v}=nev\mathbf{u}_z, $$

and hence the flow of j, through the circle whose border is the circumference, is

$$ \int {\mathbf{j.}d\mathbf{S}} =j\pi r^{2}=nev\pi r^{2}. $$

Applying Ampère’s law to the circumference gives

$$ B2\pi r=\mu _0 nev\pi r^{2}\quad \Rightarrow \quad B=\frac{\mu _0 nevr}{2}\,. $$

(c) The outward radial component of the net force on one of charges q is

$$ F=qE-qvB=q\frac{ner}{2\varepsilon _0 }-qv\frac{\mu _0 nevr}{2}=\frac{ne^{2}r}{2}\left( {\frac{1}{\varepsilon _0 }-\mu _0 v^{2}} \right) . $$

11.14. :

There are two parallel conducting plates , one of which, called the anode, has an orifice, and a potential $V=$20 V with respect to the other plate, called the cathode. By illuminating the cathode, electrons of electrical charge $q=-e=-1.602\times$ 10$^{-19}$ C and of mass $m=9.107\times$ 10$^{-31}$ kg, can be extracted from this plate via the photoelectric effect, thereby leaving it with negligible velocity. Calculate the velocity of the electrons that pass through the orifice of the anode.

Solution

Suppose that the distance between the plates is d. With this distance and the difference of potential V,  the electric field E can be calculated. With the data q, E and m the acceleration, the velocity and the displacement towards the anode can be calculated successively. In effect

$$ E=\frac{V}{d}\quad \Rightarrow \quad a=\frac{F}{m}=\frac{qE}{m}=\frac{eV}{md}\quad \Leftrightarrow \quad \frac{dv}{dt}=\frac{eV}{md}\quad \Rightarrow \quad v=\frac{eV}{md}t\,, $$

and since the distance travelled from leaving the cathode can be calculated from the velocity, then

$$ \frac{dx}{dt}=\frac{eV}{md}t\quad \Rightarrow \quad d\equiv x=\frac{eV}{2md}t^{2}\quad \Rightarrow \quad t=d\sqrt{\frac{2m}{eV}}\,. $$

Substitution of this time into the previous equation finally gives

$$ v=\frac{eV}{md}t=\frac{eV}{md}d\sqrt{\frac{2m}{eV}}=\sqrt{\frac{2eV}{m}}\,. $$

In the resolution of this problem, the use of the principle of conservation of energy is advantageous (the sum of the kinetic and potential energies when leaving the cathode is the same as that when arriving at the plane of the anode), in fact,

$$\begin{aligned} 0+qV_c =\frac{1}{2}mv^{2}+qV_a \quad \Rightarrow \quad -eV_c =\frac{1}{2}mv^{2}-eV_a \quad \Rightarrow \quad v&=\sqrt{\frac{2e(V_a -V_c )}{m}}\\&=\sqrt{\frac{2eV}{m}}. \end{aligned}$$

11.15. :

A free particle, of positive charge q and mass m, penetrates into a region of space where a homogenous and stationary magnetic field B is present. The velocity $\mathbf{v}_{0}$ of the particle when entering this region forms an angle $\alpha$ with field B. (a) Explain the type of trajectory that the particle will follow in the magnetic field for $\alpha =0$ and for $\alpha =90^{\circ }$. (b) If, in addition to B with $\alpha =90^{\circ }$, there is also a homogeneous and stationary electric field E, what would the direction of E have to be to render the movement of the particle rectilinear and uniform?

Fig. 11.19

Initial velocity

Solution

• (a) The applied force is

$$ \mathbf{F}=q\mathbf{v}\times \mathbf{B}, $$

which for $\alpha =0$ gives

$$ F=qvB\mathrm{sin} 0=0, $$

and since the fundamental equation of dynamics demands that the acceleration is null, therefore the velocity is constant and the trajectory is a straight line and parallel to B.

However for $\alpha =90^{\circ }$ the modulus of the force is

$$ F=qvB \mathrm{sin}\, 90^{\circ }=qv\mathrm{B}, $$

and the direction is obtained simply by observing Fig. 11.19. A positive charge of initial velocity $\mathbf{v}_{0}$, undergoes an increase of velocity dv in a very short time dt. Since the force is perpendicular to $\mathbf{v}_{0}$ and B, hence the acceleration $\mathbf{a}=d\mathbf{v}/dt$, and dv are also perpendicular. Therefore, the new velocity will be in the plane perpendicular to B and the charge will follow in the plane perpendicular to B.

(b) If the resultant force of both fields must be null, then the following must happen

$$ \mathbf{F}=q\mathbf{E}+q\mathbf{v}\times \mathbf{B}=0. $$

Therefore the vectors E and v $\times$ B must be of equal modulus and opposed directions, that is, E must be of direction opposite to that of drawn force F and of modulus $E=vB$.

11.16. :

A magnetic field $\mathbf{B}=B \mathbf{u}_{z}$ is homogeneous and stationary. An electron of mass m and charge $q=-e$ is impelled from the origin of coordinates with the velocity $\mathbf{v}_{0}=v_{0y}{} \mathbf{u}_{y}+v_{0z}{} \mathbf{u}_{z}$ (Fig. 11.20). (a) Calculate the radius of the helix that it describes, the pitch OP of the helix,  and the coordinates of the point P after the first revolution. (b) If the angle of $\mathbf{v}_{0}$ with OZ is small, calculate the position of P.

Fig. 11.20

Trajectory

Solution

(a) For this problem the time taken for one turn is obtained from (11.15)

$$ T=\frac{2\pi m}{qB}\,. $$

As the component of the velocity parallel to B is $v_{0z}$ and remains constant, the distance travelled in the direction of B in the time T in which the particle completes a turn is

$$ OP=v_{0z} T=\frac{2\pi m}{qB}v_{0z}\,. $$

Therefore, point P has coordinates (0, 0, $\dfrac{2\pi m}{qB}v_{0z} )$.

(b) The component of the velocity on the direction of B is $v_{0z}=v_{0}\mathrm{cos}\theta$, where $v_{0}$ is the modulus of the velocity and $\theta$ is the angle that $\mathbf{v}_{0}$ forms with B. If the angle is small, a series expansion of cos $\theta$ allows the following to be written

$$ v_{0z} =v_0 \cos \theta \approx v_0 \left( {1-\frac{\theta ^{2}}{2}+\cdots } \right) \approx v_0. $$

Therefore point P is located in the position (0, 0, $\frac{2\pi m}{qB}v_0 )$ for any small value of $\theta$. That is, if a jet of electrons leaves O, each one in a different direction (but directions close to that of B), all are focused on the same point. This fact forms the foundation of certain electron microscopes.

11.17. :

A current j circulates through a long conductive sheet of width $w=15$ mm and thickness $d=3$ mm. The sheet is introduced into a magnetic field $B=2$ T perpendicular to the plane of the sheet. The difference of potential between the edge towards which the product $\mathbf{B}\times \mathbf{j}$ points and the opposite edge is $V_{H}=3\,\upmu \mathrm{V}$. The mass density of the sheet is $\rho =8000\,\mathrm{kg/m}^{3}$ and consists of atoms of mass $m= 1.7 \times 10^{-25}\,\mathrm{kg}$. Each atom contributes a charge carrier of unknown sign whose absolute value $1.602\times 10^{-19}\,\mathrm{C}$. Calculate: (a) The velocity of the charge carriers, (b) the intensity of the current in the circuit.

Fig. 11.21

Conductive sheet

Solution

Suppose that the current density j is towards the left. If the transported charges were positive, they would move towards the left and the force that the magnetic field would apply would be downwards; there would be an accumulation of positive charges below and the potential of the lower edge would be positive with respect to the upper edge: this is a contradiction. Therefore, the charge carriers are negative and, as j is towards the left, they move towards the right.

(a) By applying (11.17), it is deduced that the modulus of the velocity is

$$ v=\frac{V_H }{wB}=\frac{3\times 10^{-6}}{1.5\times 10^{-2}\times 2}\frac{\text {m}}{\text {s}}=10^{-4}\frac{\text {m}}{\text {s}}\,. $$

(b) The density of mass can be obtained by multiplying the mass of each atom by the number of atoms in a unit volume, therefore

$$ \rho =nm\quad \Rightarrow \quad n=\frac{\rho }{m}\,. $$

Fig. 11.21 shows that, over time dt, the charges move the distance vdt, and therefore all the charges contained in the parallelepiped of volume wdv.dt, which are $dq=nqwdv.dt$, cross its face on the right. The current intensity through this face is

$$\begin{aligned} I=\frac{dq}{dt}=\frac{\rho qwdv}{m}&=\frac{8000\times 1.602\times 10^{-19}\times 1.5\times 10^{-2}\times 3\times 10^{-3}\times 10^{-4}}{{1.7\times 10}^{-25}}\text {A}\\&=33.92\;\text {A}. \end{aligned}$$

11.18. :

Figure 11.22 shows a device that corresponds to a curved tube , of average radius R, with two slots $S_{1}$ and $S_{2}$. Within the device there is a homogeneous field B pointing out of the paper. A group of particles are introduced through slot $S_{1}$ in the direction drawn, all of mass m and charge q, but which differ in the modulus of velocity. Determine: (a) Which particles pass through the $S_{2}$ slot. (b) The time taken for particles to travel from $S_{1}$ to $S_{2}.$ (c) If the particles that leave $S_{2}$ continue to be subjected to the same field B, determine the modulus and direction of a homogeneous and stationary field E outside the tube, such that the movement of particles upon leaving the tube is rectilinear and uniform.

Fig. 11.22

Tube

Solution

The trajectory followed within the tube is circular, as drawn, and implies that the acceleration is directed towards its centre of curvature, and since the force q v $\times$ B has the same direction as the acceleration, it follows that charge q is positive.

(a) The projection of the fundamental equation of dynamics in the radial direction and towards the centre of curvature gives

$$ qvB=m\frac{v^{2}}{R}\quad \Rightarrow \quad v=\frac{qBR}{m}\,, $$

which provides the dependency of v on the data of the problem and therefore it is the velocity of the particle arriving at slot S2.

(b) The length of the trajectory is 2$\uppi R/4$ and the velocity has already been calculated, therefore the time taken for the route is

$$ t=\frac{l}{v}=\frac{2\pi R/4}{qBR/m}=\frac{\pi m}{2qB}\,. $$

(c) For the velocity to be constant on the trajectory once having left the tube, the acceleration must be null, and hence the force of the electromagnetic field must be zero, that is, at any point of the rectilinear trajectory

$$ q\mathbf{E}+q\mathbf{v}\times \mathbf{B}=0. $$

Since v $\times$ B is towards the left, E must be towards the right. Otherwise, the projection of the last equation towards the left is

$$ qE+qvB\text {sen}90^{\circ }=0\quad \Rightarrow \quad E=-vB=-\frac{qB^{2}R}{m}. $$

Problems C

11.19. :

A beam of particles of charge q enters Aston’s mass spectrometer . This apparatus is formed (see the Fig. 11.23) by a section of a cylindrical capacitor of smaller radius $R_{1}$ and greater radius $R_{2}$ and an aperture S. The particles that pass through S after following the arc of average radius R, perpendicularly penetrate a cylindrical region of radius $R',$ where there is a magnetic field B inside the cylinder, perpendicular to the plane of the drawing. When leaving the field there is a detector D. The difference of potential between the armature of the capacitor is V. Calculate: (a) The electric field at the points of the arc of radius R inside the capacitor. (b) The velocity of the particles that pass through S. (c) The mass m of those particles which leave towards the detector in a direction perpendicular to that of entry into the magnetic field.

Fig. 11.23

Mass spectrometer

Solution

(a) In order to calculate the electric field, a closed surface is drawn formed by a portion of a cylindrical surface of radius r and length L located between the plates and the rest of the surface enclosing the lower plate. Discounting the effect of the edges, that is, considering that the electric field is null except in the space between the armatures, Gauss’s theorem establishes that

$$ r\theta LE=\frac{q_i }{\varepsilon _0 }\quad \Rightarrow \quad E=\frac{q_i }{\varepsilon _0 r\theta L}\,, $$

where $\theta$ is the angle formed by the drawn radii $R_{1}$ and $R_{2}$, and the electrical charge $q_{i}$ is on the lower plate.

In order to relate the electric field to the difference of potential between the upper and lower armatures, the following are used

$$ \begin{array}{l} \displaystyle { V=\int _{R_1 }^{R_2 } {Edr=} \int _{R_1 }^{R_2 } {Edr=} \int _{R_1 }^{R_2 } {\frac{q_i }{\varepsilon _0 r\theta L}dr=\frac{q_i }{\varepsilon _0 \theta L}} \ln \frac{R_2 }{R_1 }} \\ \displaystyle { \Rightarrow \quad V=ER\ln \frac{R_2 }{R_1 }\quad \forall r=R }\\ \displaystyle { \Rightarrow \quad E=\frac{V}{R\ln (R_2 /R_1 )}=\frac{V}{(R_1 +R_2 )/2\times \ln (R_2 /R_1 )} }\,.\\ \end{array} $$

(b) The particles that pass through S are those that follow the arc of radius R. Since the magnetic field is perpendicular to this trajectory, the component of that of the fundamental equation of dynamics in the direction perpendicular to the trajectory is

$$ qE=m\frac{v^{2}}{R}\quad \Rightarrow \quad v^{2}=\frac{qRE}{m}=\frac{qRV}{m(R_1 +R_2 )/2\times \ln (R_2 /R_1 )}\,. $$

Therefore only those particles whose square of their velocity satisfies this equation pass through S, thereby producing a particle selection.

(c) From Fig. 11.23, it is deduced that the radius of the arc travelled inside the magnetic field is $R'.$ The perpendicular component of the fundamental equation of dynamics gives

$$ qvB=m\frac{v^{2}}{R^{\prime }}\quad \Rightarrow \quad q^{2}B^{2}=\frac{m^{2}}{R^{\prime 2}}v^{2}\,. $$

By substituting the square of the velocity obtained in the previous section yields

$$ m=\frac{qB^{2}R'^{2}(R_1 +R_2 )\ln (R_2 /R_1 )}{2RV}\,. $$

The particles with masses different from this are not detected by D.

11.20. :

A magnetic field B acts in a cyclotron of radius R. A deuteron (formed by a proton and a neutron) is impelled between its D’s with an initial velocity $\mathbf{v}_{0}$. A difference of alternating potential of type $V=V_{0}\mathrm{cos}(\omega t)$ is established. Given that the mass and the charge of the deuteron are m and q, respectively, obtain by reasoning: (a) The velocity of the deuteron on completing the first cycle. (b) The energy of the deuteron when exiting the cyclotron. (c) The relationship that must be fulfilled between B, m, q, R and the initial velocity $v_{0}$ such that the particle makes at least one cycle. (d) What forces act in the cyclotron? Are they conservative? Explain it.

Solution

(a) The increase of kinetic energy is not due to the magnetic field but to the electric field between the D’s. In a cycle of the proton it passes twice between the D’s, therefore it absorbs energy 2q$V_{0}$. The increase of kinetic energy in the first cycle is

$$ \frac{1}{2}mv^{2}-\frac{1}{2}mv_0 ^{2}=2qV_0 \quad \Rightarrow \quad v=\sqrt{v_0 ^{2}+\frac{4qv_0 }{m}}\,. $$

This is the expression of the velocity on completing the first revolution.

(b) In the last cycle, the radius of curvature is R, and therefore the fundamental equation of dynamics for the final cycle gives

$$ qv_f B=m\frac{v_f ^{2}}{R}\quad \Rightarrow \quad v_f =\frac{qBR}{m}\quad \Rightarrow \quad E_{kf} =\frac{1}{2}mv_f ^{2}=\frac{q^{2}B^{2}R^{2}}{2m}\,. $$

(c) If N cycles were covered, the variation of the kinetic energy would be

$$ \frac{1}{2}mv_f ^{2}-\frac{1}{2}mv_0 ^{2}=N2qV_0 \quad \Rightarrow \quad N=\frac{mv_f ^{2}-mv_0 ^{2}}{4qV_0 }\,. $$

Since $N\ge 1$ must be true,

$$ mv_f ^{2}-mv_0 ^{2}\ge 4qV_0 \quad \Rightarrow \quad m\left( {\frac{qBR}{m}} \right) ^{2}-mv_0 ^{2}\ge 4qV_0 \quad \Rightarrow \quad \frac{q^{2}B^{2}R^{2}}{m}\ge mv_0 ^{2}+4qV_0\,. $$

d) The force of magnetic fields acts everywhere. The force of the electric acts in between the D’s.

The work of the forces along a closed line is

$$ \oint {\mathbf{F}.d\mathbf{l}=} \oint {q\mathbf{E}.d\mathbf{l}+\oint {q\left( {\mathbf{v}\times \mathbf{B}} \right) .d\mathbf{l}=}\,qV_0 +qV_0 +0=}\,2qV_0\,. $$

Magnetic force is not conservative. Electric force is conservative.

11.21. :

A betatron has an electromagnet whose polar pieces produce a magnetic field $\mathbf{B}_{1}$ within a circle of radius $R_{1,}$ and produce a magnetic field $\mathbf{B}_{2}$ within an annulus concentric with the circle and of inner radius $R_{1}$ and outer radius $R_{2} =2 R_{1}$. A proton is introduced so that it reaches energy $E_{k}$ following the orbit of radius $R_{2}$. (a) Calculate the average magnetic field in the circle of radius $R_{2}$, assuming $B_{1}$ and $B_{2 }$are known. (b) Assuming that the magnetic fields $\mathbf{B}_{1}$ and $\mathbf{B}_{2}$ are unknown, calculate them.

Solution

(a) The magnetic field averaged over the area of the circle bounded by the orbit is

$$\begin{aligned} B_m =\frac{1}{S}\int _S {BdS=} \frac{1}{S}\left( {\int _{S_1 } {B_1 dS_1 } +\int _{S_{2-1} } {B_2 dS_{2-1} } } \right)&=\frac{B_1 \pi R_2^2 /4+B_2 \pi \left( {R_2^2 -R_2^2 /4} \right) }{\pi R_2^2 }\\&=\frac{1}{4}B_1 +\frac{3}{4}B_2\,. \end{aligned}$$

(b) The charged particle circulates in the orbit of constant radius if the magnetic field on the orbit is half that of the average magnetic field , $B_{2}=B_{m}$/2, and hence, taking into account the value calculated for $B_{m,}$ gives

$$ 2B_2 =\frac{1}{4}B_1 +\frac{3}{4}B_2 \quad \Rightarrow \quad B_1 =5B_2\,. $$

Furthermore, the fundamental equation of dynamics gives a perpendicular component

$$ qvB_2 =m\frac{v^{2}}{R_2 }\quad \Rightarrow \quad v=\frac{qR_2 B_2 }{m}\,. $$

The kinetic energy is related to the velocity by $E_{k}=mv^{2}/2$ and, considering the previous equality, gives

$$ E_k =\frac{m}{2}\left( {\frac{qR_2 B_2 }{m}} \right) ^{2}\quad \Rightarrow \quad B_2 =\frac{\left( {2mE_k } \right) ^{1/2}}{qR_2 }\quad \Rightarrow \quad B_1 =5\frac{\left( {2mE_k } \right) ^{1/2}}{qR_2 }\,. $$

11.22. :

A force of constant direction and modulus $F=$10000 N is applied to a stone block of mass $m=2$ kg at rest. Calculate the velocity of the block after time $t=20$ h. Comment the results.

Solution

Take the OZ axis as the direction of the applied force. With this choice, the components of the force are: $F_{x}= F_{y}=0$, $F_{z}=F$.

Application of the fundamental equation of classical dynamics for the components of the acceleration gives: $a_{x}=\mathrm{a}_{y}=0$, $a_{z}=F_{z}/m=F/m=10000/2\,\mathrm{m/s}^{2}=5000\,\mathrm{m/s}^{2}$.

Therefore, from v $=\smallint$ adt, the acquired velocity has the components: $v_{x}=0$, $v_{y}=0$, $v_{z}=a_{z}t=5000 \times 20 \times 60 \times 60\,\mathrm{m/s}=3.6 \times 10^{8}\,\mathrm{m/s}$.

As this calculated velocity is greater than the speed of light, this value is impossible. Thus a relativistic approach is necessary.

From the fundamental equation of relativistic dynamics , (11.28), the three components are:

$$ 0=\frac{d}{dt}\left( {\frac{mv_x }{\sqrt{1-v^{2}/c^{2}}}} \right) \quad \Rightarrow \quad \frac{mv_x }{\sqrt{1-v^{2}/c^{2}}}=k_1 =\frac{0m}{\sqrt{1-0^{2}/c^{2}}}=0\quad \Rightarrow \quad v_x =0\,. $$

Idem $v_{y}=0$

$$ \begin{array}{l} \displaystyle { F=\frac{d}{dt}\left( {\frac{mv_z }{\sqrt{1-v^{2}/c^{2}}}} \right) =\frac{d}{dt}\left( {\frac{mv_z }{\sqrt{1-v_z ^{2}/c^{2}}}} \right) \quad \Rightarrow \quad \frac{mv_z }{\sqrt{1-v_z ^{2}/c^{2}}}=Ft+k_2 =Ft} \\ \displaystyle { \Rightarrow \frac{m^{2}v_z ^{2}}{1-v_z ^{2}/c^{2}}=F^{2}t^{2}\quad \Rightarrow \quad v_z =\left( {\frac{1}{\frac{1}{c^{2}}+\frac{m^{2}}{F^{2}t^{2}}}} \right) ^{1/2}} \\ \displaystyle { =\left( {\frac{1}{\frac{1}{299792458^{2}}+\frac{2^{2}}{10000^{2}(20\times 60\times 60)^{2}}}} \right) ^{1/2}\text {m/s}=2.3037\times 10^{8}\text {m/s}}\,. \end{array} $$

11.23. :

In a region of space there is an electromagnetic field that, with respect to a reference system, has the components $E_{x}=0$, $E_{y}=3$ V/m, $E_{z}=0$, $B_{x}=0.2$ T, $B_{y}=0$, and $B_{z}=0$. Express the equations of motion of a particle of mass m and charge q with a high velocity.

Solution

The force that the field applies on the charge is

$$ \mathbf{F}= q(\mathbf{E+v}\times \mathbf{B})=q(3\mathbf{u}_y +\left| {{\begin{array}{llll} {\mathbf{u}_x }&{} {\mathbf{u}_y }&{} {\mathbf{u}_z} \\ {v_x }&{} {v_y }&{} {v_z } \\ {0.2}&{} 0&{} 0 \end{array} }} \right| )=q\left[ {\left( {3+0.2v_z } \right) \mathbf{u}_y -0.2v_y \mathbf{u}_z } \right] . $$

If the velocity is high, the laws of relativistic dynamics (11.33) must be applied, whose three components are:

$$ \begin{array}{l} \displaystyle { 0=\frac{d}{dt}\left( {\frac{mv_x }{\sqrt{1-\left( {v_x ^{2}+v_y ^{2}+v_z ^{2}} \right) /c^{2}}}} \right) } \\ \displaystyle { \Rightarrow \quad \frac{mv_x }{\sqrt{1-\left( {v_x ^{2}+v_y ^{2}+v_z ^{2}} \right) /c^{2}}}=k_1 =\frac{mv_{x0} }{\sqrt{1-\left( {v_{x0} ^{2}+v_{y0} ^{2}+v_{z0} ^{2}} \right) /c^{2}}}}\,, \\ \end{array} $$

$$ \begin{array}{l} \displaystyle { {q}\left( {3-0.2v_z } \right) =\frac{d}{dt}\left( {\frac{mv_y }{\sqrt{1-\left( {v_x ^{2}+v_y ^{2}+v_z ^{2}} \right) /c^{2}}}} \right) } \\ \displaystyle { \Rightarrow \quad \frac{mv_y }{\sqrt{1-\left( {v_x ^{2}+v_y ^{2}+v_z ^{2}} \right) /c^{2}}}={q}\left( {3-0.2v_z } \right) t+k_2 ={q}\left( {3-0.2v_z } \right) t+\frac{mv_{y0} }{\sqrt{1-\left( {v_{x0} ^{2}+v_{y0} ^{2}+v_{z0} ^{2}} \right) /c^{2}}}}\,, \\ \end{array} $$

$$\begin{aligned}&-0.2{q}v_{y} = \frac{d}{dt} \left( \frac{mv_{z}}{\sqrt{1-(v_{x}^{2} + v_{y}^{2} + v_{z}^{2}/c^{2})}}\right) \Rightarrow \frac{mv_{z}}{\sqrt{1-(v_{x}^{2} + v_{y}^{2} + v_{z}^{2}/c^{2})}}\\&=-0.2{q}v_{y}t + k_{3} = -0.2{q}v_{y}t + \frac{mv_{z0}}{\sqrt{1-(v_{x0}^{2} + v_{y0}^{2} + v_{z0}^{2}/c^{2})}}\,. \end{aligned}$$

11.24. :

A cyclotron has a radius $R=4$ m and a magnetic field $B=0.012$ T. (a) Find the velocity of the electrons whose charge is of absolute value $e=1.602\times$10$^{-19}$ C and whose mass is $m=9.107\times$10$^{-31}$ kg. (b) For the protons of equal charge to that of the electrons except for the sign and the mass $m_{p}=1.673 \times 10^{-27}$ kg, calculate the velocity that they acquire.

Solution

Fundamental equation of classical dynamics

$$ \mathbf{F}=m\mathbf{a}\quad \Rightarrow \quad q\mathbf{v}\times \mathbf{B}=m\mathbf{a}. $$

In absolute values

$$ qvB=mv^{2}/R\quad \Rightarrow \quad v=qBR/m. $$

For the electrons, we have

$$ v=\frac{1.602\times 10^{-19}\times 0.012\times 4}{9.107\times 10^{-31}}\text {m/s}=8.444\times 10^{8}\text {m/s}. $$

This result is impossible since no object carrying energy can move faster than light in a vacuum. Therefore the fundamental equation of dynamics used is inapplicable in this case.

For the protons

$$ v=\frac{1.602 \times 10^{-19}\times 0.012\times 4}{1.673 \times 10^{-27}}\text {m/s}=4.596\times 10^{5}\text {m/s}. $$

This result is acceptable since protons have a much greater mass than the electrons but have equal energy, and hence their velocity is much smaller.

If the fundamental equation of relativistic dynamics is applied to the electrons, it is sufficient to substitute the value of the energy $E_{r}$, (11.29), into the expression of the radius of curvature, (11.52), to obtain

$$ \begin{array}{l} \displaystyle { v=\left( {\frac{R^{2}e^{2}B^{2}}{m^{2}+R^{2}e^{2}B^{2}/c^{2}}} \right) ^{1/2}} \\ \displaystyle { =\left( {\frac{4^{2}\times 1.602^{2}\times 10^{-38}\times 0.012^{2}}{9.107^{2}\times 10^{-62}+4^{2}\times 1.602^{2}\times 10^{-38}\times 0.012^{2}/\left( {2.99792458^{2}\times 10^{16}} \right) }} \right) ^{1/2}\text {m/s}} \\ \displaystyle { =2.7924\times 10^{8}\text {m/s} }. \end{array} $$

Observe that the result for the electron is very different from that previous and constitutes the only good result.

11.25. :

Spain’s new 3-GeV synchrotron , Alba (Spanish for “dawn light”), appeared on-line in 2010. The first seven beamlines are a mixture of soft X rays, (for applications in material science, solid-state physics, biology , chemistry , and medicine ), and of hard X rays for crystallography and absorption studies. The electrons acquire energy of 3 GeV in the accelerator of the machine and enter the storage ring. It is supposed that upon passing through a small zone in this ring where there is a magnetic field of 10 T, the electrons are accelerated by being deflected 3$^{\circ }$ from their straight trajectory, in order to emit X-rays. Given that the mass of the electron is $m=9.109\times 10^{-31}$ kg and its electrical charge is negative and of absolute value $e=1.602\times 10^{-9}$ C, and Planck’s constant is $h=6.626\times 10^{-34 }$Js $=4.136\times 10^{-15}$ eVs, calculate: (a) the velocity of the electrons; (b) the time they remain within the magnetic field; (c) the radius of curvature in the magnetic field of 10 T; (d) If the energy of the emitted X-rays is 10 keV, calculate their frequency and wavelength. (1 eV $=1.602\times 10^{-19}$ C $\times$1V $=1.602\times 10^{-19}$ J).

Solution

(a) From (11.31), the following is given

$$ v=c\sqrt{1-\left( {\frac{9.109\times 10^{-31}\times 299792458^{2}}{3\times 10^{9}\times 1.602\times 10^{-19}}} \right) ^{2}}=0.999999971c=299792449\;\text {m/s}, $$

therefore, the electrons are relativistic.

(b) If the magnetic field of 10 T were unlimited in space, then the period that the protons would remain within the magnetic field would be

$$ T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}=\frac{2\pi \times 9.109\times 10^{-31}}{1.602\times 10^{-19} \times 10}\text {s}=357.3\times 10^{-14}\text {s}. $$

Hence the time spent during the rotation of 3$^{\circ }$ is

$$ t=\frac{3^{\circ }}{360^{\circ }}\frac{2\pi m}{qB}=\frac{3}{360}\frac{2\pi 9.109\times 10^{-31}}{1.602\times 10^{-19}\times 10}\text {s}=2.978\times 10^{-14}\text {s}. $$

(c) The radius is calculated from (11.52), giving

$$ R=\frac{vE_r }{\text {q}c^{2}B}\approx \frac{E_r }{\text {q}cB}\approx \frac{3\times 1.602\times 10^{-19}\times 10^{9}}{1.602 \times 10^{-19}\times 3\times 10^{8}\times 10}=1.000\;\text {m}. $$

(d)

$$ E_r =h\nu \Rightarrow \nu =\frac{E_r }{h}=\frac{10000\times 1.602\times 10^{-19}}{6.626\times 10^{-34}}\text {Hz}\,=\,{2.4177}\times 10^{18}\text {Hz}. $$

Therefore, the wavelength is

$$ \lambda =\frac{c}{\nu }=\frac{299792458}{2.4177\times 10^{18}}\;\text {m}\,=\,{1.2408}\times {10}^{-10}\;\text {m}=0.12408\;\text {nm}, $$

which is of the size of the atoms.