IB Physics HL D.3 Motion in Electromagnetic Fields Question Bank

Practise IB Physics HL D.3 by solving charged-particle motion through combined electric and magnetic fields and evaluating trajectory data.

Exam points

Combine electric, magnetic and centripetal forces to model helical or curved charged-particle paths.
Relate field work, potential and kinetic-energy changes in multi-stage particle motion.
Evaluate velocity selectors, mass-spectrometer-style evidence and uncertainty in HL data.

Covered syllabus points

D.3.1—Charge in electric fields D.3.2—Charge in magnetic fields D.3.3—Crossed electric and magnetic fields D.3.4—Magnetic force on charge D.3.5—Force on current conductor D.3.6—Force between parallel wires

[Maximum number: 4]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

(a)

(b) The diagram shows a pair of horizontal metal plates. Electrons can be deflected vertically using an electric field between the plates.

[ 1 ]

(i)

Label, on the diagram, the polarity of the metal plates which would cause an electron positioned between the plates to accelerate upwards.

[ 1 ]

(b)

The diagram shows two isolated electrons, X and Y , initially at rest in a vacuum. The initial separation of the electrons is 5.0 mm . The electrons subsequently move apart in the directions shown.

[ 3 ]

(i)

Discuss the motion of one electron after it begins to move.

[ 3 ]

[Maximum number: 8]

A rod, R , lies perpendicular to a uniform magnetic field B of strength 0.50 T directed into the plane of the page. R is connected to a circuit and the electric current $I_{\mathrm{R}}$ is 2.0 A .

(a)

Calculate the magnitude of the force per unit length on R.

[ 2 ]

$\begin{aligned} \text { Use of } F=\text { BIL OR F/L = BI } =1.0 \text { « } \mathrm{N} \mathrm{~m}^{-1} \text { » } \text { } \end{aligned}$

(b)

Identify on the diagram the direction of the force on R.

[ 1 ]

(c)

An electron is moving at constant velocity $v=9.0 \times 10^{5} \mathrm{~ms}^{-1}$ parallel to a rod T , at a perpendicular distance d. T is connected to a cell via a switch. When the switch is closed the electron's path curves with initial radius r. The magnetic field strength at the position of the electron at d is $4.0 \times 10^{-6} \mathrm{~T}$ .

[ 5 ]

(i)

Calculate the initial radius of the electron's path.

[ 3 ]

use of $\mathrm{qvB}=\mathrm{mv}^{2} / \mathrm{r}$

OR
$\begin{aligned} & r=\frac{m v}{q B} & 1.6 \times 10^{-19} \times 4.0 \times 10^{-6}=\frac{9.11 \times 10^{-31} \times 9 \times 10^{5}}{r} \end{aligned}$

OR
$\begin{aligned} & r=\frac{\left(9.11 x 10^{-31}\right)\left(9 x 10^{5}\right)}{\left[\left(1.6 \times 10^{-19}\right)\left(4 x 10^{-6}\right)\right]} & r=1.28 \text { OR } 1.3 \text { «m» } \end{aligned}$

(ii)

Explain the subsequent change in the radius of the electron's path.

[ 2 ]

[Maximum number: 5]

Two oppositely charged parallel plates are a distance 8.0 cm apart. The potential difference between the plates is 120 V . An alpha particle is placed on the positively charged plate and released from rest. Gravity is ignored.

Figure for Question D.3 Motion in electromagnetic fields question 3 — IB Physics HL

(a)

Show that the acceleration of the alpha particle is about $7 \times 10^{10} \mathrm{~ms}^{-2}$ .

[ 2 ]

$\begin{aligned} F=<e E=>3.2 \times 10^{-19} \times 1.5 \times 10^{3} \text { OR } 4.8 \times 10^{-16} \mathrm{~N} a=<\frac{F}{m}=>\frac{4.8 \times 10^{-16}}{4 \times 1.67 \times 10^{-27}} \text { OR } 7.2 \times 10^{10} \mathrm{~ms}^{-2} \quad \end{aligned}$

Allow ECF from a).

Award [1] if they use a charge of e and a mass of 2 u obtaining the right result.

(b)

A magnetic field directed into the plane of the page is now established between the plates. An alpha particle enters the region between the plates with a horizontal speed of $5.0 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}$ . The particle is not deflected.

Calculate the magnitude of the magnetic field.

[ 2 ]

$\begin{aligned} F_{\mathrm{m}} =F_{\mathrm{e}} \text { OR } q v B=q E \text { OR } B=F_{\mathrm{e}} / q v \quad B =\ll \frac{E}{v}=\frac{1.5 \times 10^{3}}{5.0 \times 10^{5}}=>3.0 \times 10^{-3} T \end{aligned}$

Award [2] if $3.0 \times 10^{-3}$ « T » is seen as the answer without working.

Allow ECF from a) and b) i)

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(c)

The alpha particle in (c) is replaced by an electron. The electron enters the region between the plates with the same velocity as the alpha particle.

Draw, on the diagram, the path of the electron.

[ 1 ]

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Covered syllabus points

D.3 Motion in electromagnetic fields question 1

Question (a)

Question (i)

Question (b)

Question (i)

D.3 Motion in electromagnetic fields question 2

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

D.3 Motion in electromagnetic fields question 3

Question (a)

Question (b)

Question (c)