IB Physics SL D.3 Motion in Electromagnetic Fields

[Maximum number: 4]

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.

(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.

[2]

(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)

[2]

[Maximum number: 4]

(a)

A proton moves on a circular path in a region of uniform magnetic field of magnetic flux density B that is directed into the plane of the page.

[ 4 ]

(i)

On the diagram, draw an arrow to indicate the velocity of the proton at the position shown.

[ 1 ]

(ii)

Show that the frequency of revolution of the proton is given by $f=\frac{e B}{2 \pi m_{\mathrm{p}}}$ .

[ 3 ]

$\begin{aligned} e v B=\frac{m_{\mathrm{p}} v^{2}}{R} f=\frac{1}{T}=\frac{v}{2 \pi R} \end{aligned}$

Both q and e are acceptable for the
charge.

[3]

Algebra leading to required expression $\boldsymbol{<}=\frac{e B}{2 \pi m_{\mathrm{p}}} \boldsymbol{} \boldsymbol{}$

[Maximum number: 2]

In an experiment a student constructs the circuit shown in the diagram. The ammeter and the voltmeter are assumed to be ideal.

The diagram shows a cross-sectional view of the connecting wire in (c).

The wire which carries a current of 3.5 A into the page, is placed in a region of uniform magnetic field of flux density 0.25 T . The field is directed at right angles to the wire.

Determine the magnitude and direction of the magnetic force on one of the charge carriers in the wire.

$F=<q v B=1.6 \times 10^{-19} \times 2.0 \times 10^{-4} \times 0.25=>8.1 \times 10^{-24} \mathrm{~N}$

directed down OR south

[Maximum number: 2]

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.

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

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

Allow ECF from a).

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

[2]

[Maximum number: 2]

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.

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.

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

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

Allow ECF from a) and b) i)

[2]

[Maximum number: 3]

The long straight conductor is formed into a coil consisting of two separate turns, X and Y . The coil hangs with its axis vertical.

Assume that the turns of the coil each behave as a long straight conductor.

The current in the coil is 15 A and the circumference of one turn is 0.48 m . In order to restore X and Y to their original separation, a mass of $2.8 \times 10^{-4} \mathrm{~kg}$ is suspended from turn Y . Estimate the magnetic field strength at X due to Y .

A3. This question is about radioactivity.

Caesium-137 ( ${ }_{55}^{137} \mathrm{Cs}$ ) is a radioactive waste product with a half-life of 30 years that is formed during the fission of uranium. Caesium-137 decays by the emission of a beta-minus $\left(\beta^{-}\right)$ particle to form a nuclide of barium (Ba).

$F=\left(0.280 \times 10^{-3} \times 9.81=\right) 2.75 \times 10^{-3} \mathrm{~N}$ ;
$B=\frac{F}{I l}$ or correct substitution $2.75 \times 10^{-3}=B \times(15) \times 0.48$ ;
$B=\left(\frac{2.75 \times 10^{-3}}{15 \times 0.48}=\right) 3.8 \times 10^{-4} \mathrm{~T} ;$

[Maximum number: 3]

An electrically heated pad is designed to keep a pet warm.

The pad is heated using a resistor that is placed inside the pad. The dimensions of the resistor are shown on the diagram. The resistor has a resistance of $4.2 \Omega$ and a total length of 1.25 m .

When there is a current in the resistor, the temperature in the pad changes from a room temperature of $20^{\circ} \mathrm{C}$ to its operating temperature at $35^{\circ} \mathrm{C}$ .

(a)

When there is a current in the resistor, magnetic forces act between the resistor strips.

For the part of the resistor labelled RS,

[ 3 ]

(i)

outline the magnetic force acting on it due to the current in PQ .

[ 1 ]

(ii)

state and explain the net magnetic force acting on it due to the currents in PQ and TU .

[ 2 ]

IB Physics SL D 3 Motion in Electromagnetic Fields Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (i)

Question (ii)

Question 3

Question 4

Question 5

Question 6

Question 7

Question (a)

Question (i)

Question (ii)