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.

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, 2

T , 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

N revolutions, 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

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 T is 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

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