Question

12
3
(a) Explain the use of a uniform electric field and a uniform magnetic field for the selection of the
velocity of a charged particle. You may draw a diagram if you wish.​

Answer 1

Answer:

Fig. 6.1

Data for the diameters of the paths are shown in Fig. 6.2.

Fig. 6.2

The ions in path B each have charge +1.6 × 10-19 C.

(i) Determine the mass, in u, of the ions in path B. [4]

(ii) Suggest and explain quantitatively a reason for the difference in radii of the paths A and B of the ions. [3]

Reference: Past Exam Paper – June 2014 Paper 42 Q6

Solution:

(a)

The electric and magnetic fields are normal to each other.

The charged particle initially enters the region at right angles to both fields.

For the charged particle to pass without deviation, its velocity should be v = E / B.

(b)

(i)

{The magnetic force provides the centripetal force

Centripetal force = Magnetic force

mv2 / r = Bqv}

m = Bqr / v

m = (640×10-3) (1.6×10-19) (12.4×10-2) / (9.6×104)

m = 6.61×10-26 kg

{1 u = 1.66×10-27 kg}

Mass, m (in u) = (6.61×10-26) / (1.66×10-27) = 40 u

(ii)

{ mv2 / r = Bqv

q / m = v / Br

The ions are of the same isotope, so they have the same mass m. B and v are also constants.

Charge q is proportional to 1 / r}

EITHER The ratio q/m is proportional to 1/r

OR Mass m is constant and charge q is proportional to 1/r.

{The radius of path A is half that of path B. Since q ∝ 1/r,}

The ratio q/m for path A is twice that for path B.

So, the ions in path A have (the same mass but) twice the charge (of the ions in path B).