An electron having charge –q and mass m is accelerated from rest in a vacuum through a potential difference V.

Question 3

An electron having charge –q and mass m is accelerated from rest in a vacuum through a potential difference V.

The electron then enters a region of uniform magnetic field of magnetic flux density B, as shown in Fig. 7.1.

Fig. 7.1

The direction of the uniform magnetic field is into the plane of the paper.

The velocity of the electron as it enters the magnetic field is normal to the magnetic field.

The radius of the circular path of the electron in the magnetic field is r.

(a) Explain why the path of the electron in the magnetic field is the arc of a circle. [3]

(b) Show that the magnitude p of the momentum of the electron as it enters the magnetic field is given by

p = √(2mqV).

[2]

(c) The potential difference V is 120 V. The radius r of the circular arc is 7.4 cm.

Determine the magnitude B of the magnetic flux density. [3]

(d) The potential difference V in (c) is increased. The magnetic flux density B remains unchanged.

By reference to the momentum of the electron, explain the effect of this increase on the

radius r of the path of the electron in the magnetic field. [2]

 [Total: 10]

Reference: Past Exam Paper – June 2017 Paper 41 & 43 Q7

Solution:

(a) The magnetic force always acts normal to the direction of motion. Since the magnitude of the force is constant, it provides the centripetal force.

(b)

{The electron is accelerated through a p.d. V. So, it has electric PE.

From the conservation of energy, as the electron moves, the increase in KE of the electron is equal to the loss in electric PE.}

increase in KE = loss in PE or ½ mv² = qV              

p = mv with algebra leading to p = √(2mqV)  

{Momentum = mv. So, make mv the subject of formula from the above equation.

½ mv² = qV

mv² = 2qV

Multiply by m on both sides,

m²v² = 2mqV

(mv)² = 2mqV

Take square root,

mv = √(2mqV)

p = √(2mqV)}

(c)

{The magnetic force provides the centripetal force.

Magnetic force = Centripetal force}

Bqv = mv² / r

mv = Bqr                     or p = Bqr

{Make mv (the momentum) the subject of formula here, and equate this to the formula for momentum shown above.}

{p = Bqr           and p = √(2mqV). So,

√(2mqV) = Bqr}

√ (2 × 9.11 × 10^-31 × 1.60 × 10^-19 × 120) = B × 1.60 × 10^-19 × 0.074

B = 5.0 × 10^-4 T                                              

(d)

{ p = √(2mqV). Increasing the p.d. V causes the momentum to be greater.}

There is a greater momentum.

Since (p = Bqr and), the radius r increases