Positive ions are travelling through a vacuum in a narrow beam. The ions enter a region of uniform magnetic field of flux density B and are deflected in a semi-circular arc, as shown in Fig. 5.1.

Question 4

Positive ions are travelling through a vacuum in a narrow beam. The ions enter a region of uniform magnetic field of flux density B and are deflected in a semi-circular arc, as shown in Fig. 5.1.

Fig. 5.1

The ions, travelling with speed 1.40 × 10⁵ m s^-1, are detected at a fixed detector when the diameter of the arc in the magnetic field is 12.8 cm.

(a) By reference to Fig. 5.1, state the direction of the magnetic field. [1]

(b) The ions have mass 20 u and charge +1.6 × 10^-19 C. Show that the magnetic flux density is 0.454 T. Explain your working. [3]

(c) Ions of mass 22 u with the same charge and speed as those in (b) are also present in the beam.

(i) On Fig. 5.1, sketch the path of these ions in the magnetic field of magnetic flux

density 0.454 T. [1]

(ii) In order to detect these ions at the fixed detector, the magnetic flux density is changed.

Calculate this new magnetic flux density. [2]

Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q5

Solution:

(a)

The field is into (the plane of) the paper.

{Fleming’s Left Hand rule: thumb: force, first finger: magnetic field and middle finger: current (direction of +ve charge)}

(b)

{The force acting on the ions causes them to move through the arc of a circle. Circular motion involves a centripetal force.}

The force due to the magnetic field provides the centripetal force.

Centripetal force = Magnetic force}

mv² / r = Bqv

{B = mv / rq

1 u = 1.66×10^-27 kg}

B = (20 × 1.66×10^-27) × (1.40×10⁵) / [(6.4×10^-2) × (1.6×10^-19)]

Magnetic flux density B = 0.454 T    

(c)

(i)

{From part (b), B = mv / rq

Radius r = mv / Bq

Since the mass m is now greater, while the charge q and speed v are the same, the radius would be greater than previously.}

Path is a semicircle with the diameter greater than 12.8 cm

(ii)

{From part (b), B = mv / rq

Since speed v and charge q are not changed,

B ∝ m / r

The detector is kept fixed (at its position). In order to detect the ions, the magnetic flux density should be changed such that the radius is the same as before. So, the radius r is also constant.

B ∝ m

B = km            where k is a constant

The magnetic field is proportional to the mass.

When mass m = 20 u, B = 0.454 T

0.454 = k × 20             ------------------------ (1)

When mass = 22 u, let the new magnetic flux density be B`

B` = k × 22                  ------------------------ (2)

Divide (2) by (1),

B` / 0.454 = 22 / 20

B` = (22/20) × 0.454 }

New magnetic flux density, B = (22/20) × 0.454

New magnetic flux density, B = 0.499 T