A particle has mass m and charge +q and is travelling with speed v through a vacuum. The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 6.1.

Question 2

A particle has mass m and charge +q and is travelling with speed v through a vacuum.

The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 6.1.

Fig. 6.1

The uniform electric field between the plates has magnitude 2.8 × 10⁴ V m^-1 and is zero outside the plates.

The particle passes between the plates and emerges beyond them, as illustrated in Fig. 6.1.

(a) Explain why the path of the particle in the electric field is not an arc of a circle. [1]

(b) A uniform magnetic field is now formed in the region between the metal plates. The

magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 6.2.

 Fig. 6.2

(i) State and explain the direction of the magnetic field. [2]

(ii) The particle has speed 4.7 × 10⁵ m s^-1.

Calculate the magnitude of the magnetic flux density.

Explain your working. [3]

(c) The particle in (b) has mass m, charge +q and speed v.

Without any further calculation, state the effect, if any, on the path of a particle that has

(i) mass m, charge –q and speed v, [1]

(ii) mass m, charge +q and speed 2v, [1]

(iii) mass 2m, charge +q and speed v. [1]

Reference: Past Exam Paper – November 2013 Paper 43 Q6

Solution:

(a) The path is not a circular arc in the field because the force on the particle acts vertically downwards.

{For circular motion, the force is towards the centre of the motion}

(b)

(i)  The direction of the magnetic field is into plane of page since the direction of the force due to the magnetic field should be opposite to that due to electric field

(ii)

{The magnitudes of the forces due to the magnetic and electric fields are equal.}

force due to magnetic field = force due to electric field

Bqv = qE

B = E / v

B = (2.8 × 10⁴) / (4.7 × 10⁵)

B = 6.0 × 10^–2 T

(c)

{Magnetic force = Bqv                        and Electric force = Eq

For the particle to pass undeviated, Bqv = Eq

We need to understand what the changes have on both forces and which force would then be more significant.}

(i) no change / not deviated

(ii) deviated upwards

(iii) no change / not deviated

When a constant braking force is applied to a vehicle moving at speed v, the distance d moved by the vehicle in coming to rest is given by the expression d = kv2

Question 3

When a constant braking force is applied to a vehicle moving at speed v, the distance d moved by the vehicle in coming to rest is given by the expression

d = kv²

where k is a constant.

When d is measured in metres and v is measured in metres per second, the constant has a value of k₁.

What is the value of the constant when the distance is measured in metres, and the speed is measured in kilometres per hour?

A 0.0772 k₁                       B 0.278 k₁                         C 3.60 k₁                            D 13.0 k₁

Reference: Past Exam Paper – June 2015 Paper 11 Q3

Solution:

Answer: A.

d = kv²

When speed v is in ‘m/s’ and distance d is in ‘m’, the value of k is ‘k₁’.

k₁ = d / v²

We want to know the value of k when v is in ‘km/h’ and d in ‘m’.

Converting m/s to km/h

1 m/s means that in 1s - - -> 1m

1 h = 3600 s

1s - - -> 1m

1h - - > 3600 m = 3.6 km

Thus, 1 m/.s is equal to 3.6 km/h. Let the new value of speed = v₂.

v₂ = 3.6 v

New value of k = d / v₂² = d / (3.6v)2 = (1 / 3.6²) (d / v²)

But d / v₂ = k₁  

New value of k = (1 / 3.6²) k₁ = 0.0772 k₁