A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 105 m s-1 in a vacuum.

Question 12

A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 10⁵ m s^-1 in a vacuum. The particle enters a uniform magnetic field of flux density 9.7 × 10^-2 T, as shown in Fig. 9.1.

Fig. 9.1

The magnetic field direction is perpendicular to the initial velocity of the particle and perpendicular to, and out of, the plane of the paper.

A uniform electric field is applied in the same region as the magnetic field so that the particle passes undeviated through the fields.

(a) State and explain the direction of the electric field. [2]

(b) Calculate the magnitude of the electric field strength.

Explain your working. [3]

(c) The electric field is now removed so that the positively-charged particle follows a curved path in the magnetic field. This path is an arc of a circle of radius 4.0 cm.

Calculate, for the particle, the ratio q/m. [3]

(d) The particle has a charge of 3e where e is the elementary charge.

(i) Use your answer in (c) to determine the mass, in u, of the particle. [2]

(ii) The particle is the nucleus of an atom. State the number of protons and the number

[Total: 11]

Reference: Past Exam Paper – March 2016 Paper 42 Q9

Solution:

(a) The direction of the force due to the electric field should be opposite to the force due to the magnetic field. So the electric field is up the page.

{From Fleming’s left hand rule, the direction of the force on the particle is downwards.  For the particle to pas undeviated, the force due to the electric field should be up. The direction of an electric field gives the direction of the force on a positive charge. So, the electric field should be upwards.}

(b)

force due to electric field = force due to magnetic field

or

Eq = Bqv

{Eliminating q, }

E = Bv                                                

E = 9.7×10^–2 × 1.6×10⁵

E = 1.6 (1.55) × 10⁴ V m^–1

(c)

{Force due to magnetic field provides the centripetal force.

Bqv = mv² / r }

q / m = v / Br                                                              

{The radius of the circle should be in metres.}

q / m = 1.6 × 10⁵ / (9.7×10^–2 × 4.0×10^–2)                  

q / m = 4.1 (4.12) × 10⁷ C kg^–1                                  

(d)

(i)

{q = 3e = 3 × 1.60×10^–19 C

q / m = 4.12 × 10⁷

m = q / 4.12 × 10⁷}

m = (3 × 1.60×10^–19) / (4.12×10⁷)     

{1 u = 1.66 × 10^–27 kg

To obtain the mass in u, we divide by 1.66 × 10^–2.}

m = 1.16×10^–26 / 1.66×10^–27

m = 7(.0) u       

(ii) 3 protons, 4 neutrons

{1 u is the mass of 1 unit / particle. 7 u means that there are 7 particles.

From the question, the charge of the particle is 3e – that is, it contains 3 protons. The rest are neutrons.}