(a) A particle has mass m, charge +q and speed v. State the magnitude and direction of the force, if any,

Q#7 (Past Exam Paper – November 2015 Paper 41 & 42 Q6)

(a) A particle has mass m, charge +q and speed v.

State the magnitude and direction of the force, if any, on the particle when the particle is

travelling along the direction of

(i) a uniform gravitational field of field strength g, [2]

(ii) a uniform magnetic field of flux density B. [1]

(b) Two charged horizontal metal plates, situated in a vacuum, produce a uniform electric field of field strength E between the plates. The field strength outside the region between the plates is zero.

The particle in (a) enters the region of the electric field at right-angles to the direction of the field, as illustrated in Fig. 6.1.

Fig. 6.1

A uniform magnetic field is to be applied in the same region as the electric field so that the particle passes undeviated through the region between the plates.

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

(ii) Derive, with explanation, the relation between the speed v and the magnitudes of the

electric field strength E and the magnetic flux density B. [3]

(c) A second particle has the same mass m and charge +q as that in (b) but its speed is 2v. This particle enters the region between the plates along the same direction as the particle in (b).

On Fig. 6.1, sketch the path of this particle in the region between the plates. [2]

Solution:

(a)

(i)

force = mg

along the direction of the field/of the motion

{The gravitational force on the particle in downwards, along the direction of the field (which is also downwards).}

(ii) no force

{The particle will only feel an electromagnetic force when the direction of motion is perpendicular (or has a component perpendicular) to the direction of the magnetic field.}

(b)

(i)

force due to E-field downwards so force due to B-field upwards

into the plane of the paper

{The direction of the electric field strength E indicates the direction of the electric force on a positive charge. So, the electric field would cause a downward force on the particle.

For the particle to pass undeviated, the force exerted by the magnetic field should be upwards.

From Fleming’s left hand rule,

Thumb = force = up

Middle finger = current = to the right

So, forefinger = (magnetic) field = into plane of paper}

(ii)

force due to magnetic field = Bqv

force due to electric field = Eq

the forces are equal (and opposite)

{Bqv = Eq

Bv = E}

so Bv = E                    

or Eq = Bqv so E = Bv

(c) sketch: smooth curved path in ‘upward’ direction

{Remember that the magnetic force is upwards while the electric force is downwards.

The speed v only affects the magnetic force (= Bqv) and NOT the electric force (= Eq; does not depend on v).

Since the speed is now doubled, the magnetic force would be greater and the particle would deviate upwards.}

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