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PHYSICS, M6 2016 HSC 23b

The diagram shows electrons travelling in a vacuum at `5.2 × 10^(4) \ text{m s}^(-1)` entering an electric field of `10 \ text {V m}^(-1)`.

 

A magnetic field is applied so that the electrons continue undeflected.

What is the magnitude and direction of the magnetic field?   (3 marks)

Show Answers Only

`0.0002  text{T Up the page.}`

Show Worked Solution

For the electrons to continue undeflected the force on the electrons due to the electric field must be equal in magnitude (but opposite in direction) to the force on them due to the magnetic field.

`F_(E)` `=F_(b)`
`qE` `=qvB`
`B` `=(E)/(v)`
  `=(10)/(5.2 xx10^(4))`
  `=0.0002  text{T (up the page)}`

PHYSICS, M6 2024 HSC 17 MC

The diagram shows a type of particle accelerator called a cyclotron.

Cyclotrons accelerate charged particles, following the path as shown.
 

   

An electric field acts on a charged particle as it moves through the gap between the dees. A strong magnetic field is also in place.

Once a charged particle has the required velocity, it exits the accelerator towards a target.

Which of the following is true about a charged particle in a cyclotron?

  1. It increases speed while inside the dees.
  2. It only accelerates while between the dees.
  3. It undergoes acceleration inside and between the dees.
  4. It slows down inside the dees and speeds up between the dees.
Show Answers Only

\(C\)

Show Worked Solution
  • When the charged particle is between the dees, it will experience an acceleration due to the electric field present where \(a = \dfrac{qE}{m}\).
  • When the charged particle is inside of the dees, the particle undergoes uniform circular motion due to the strong magnetic field in place from the electromagnets.
  • While the magnitude of the velocity of the charged particle does not change, the direction of the velocity does. Hence, there is a change in velocity of the particle so it is experiencing an acceleration.
  • A charged particle will experience a centripetal force/acceleration when moving perpendicular to a magnetic field.

\(\Rightarrow C\)

♦ Mean mark 45%.

PHYSICS, M6 2024 HSC 19 MC

In a vacuum chamber there is a uniform electric field and a uniform magnetic field.

A proton having a velocity, \(v\), enters the chamber. Its velocity remains unchanged as it travels through the chamber.

A second proton having a velocity, \(2v\), in the same direction as the first proton, then enters the chamber at the same point as the first proton.

In the chamber, the acceleration of the second proton

  1. is zero.
  2. is constant in magnitude and direction.
  3. changes in both magnitude and direction.
  4. is constant in magnitude, but not direction.
Show Answers Only

\(C\)

Show Worked Solution
  • The first proton, with velocity \(v\), travels through with no deflection. The net force acting on the proton is zero.
  • However, as the second proton has a velocity of \(2v\), \(F_B > F_E\) and the initial direction of \(F_B\) will be in the opposite direction of \(F_E\). This will cause the second proton to undergo circular motion (force is perpendicular to the velocity). 
  • The acceleration of the second proton will change direction as the centripetal force/acceleration will act towards the centre of the circular path that the second proton undertakes.
  • The magnitude of the acceleration of the second proton will also change. Initially \(F_B\) opposes \(F_E\) and \(F_{\text{net}}\) is at a minimum. As the direction of \(F_B\) changes, \(F_{\text{net}}\) will increase as \(F_E\) will not directly oppose \(F_B\).
  • As the magnitude of the net force on the second proton increases, the magnitude of the acceleration on the second proton will also increase.

\(\Rightarrow C\)

♦♦♦ Mean mark 26%.

PHYSICS, M6 2020 VCE 3

Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).

Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
 

  1. Show that the velocity of an electron that travels straight through the aperture to point \(\text{Y}\) is given by  \( v_{0} \) = \( \dfrac{E}{B}\).   (1 mark)

  2. Calculate the magnitude of the velocity, \(v_0\), of an electron that travels straight through the aperture to point \(\text{Y}\) if  \(E\) = 500 kV m\(^{-1}\)  and  \(B\) = 0.25 T. Show your working.   (2 marks)

  3.  i. At which of the points – \(\text{X, Y}\), or \(\text{Z}\) – in Figure 2 could electrons travelling faster than \(v_0\) arrive?   (1 mark)

  4. ii. Explain your answer to part c.i.   (2 marks)

Show Answers Only

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  

 
b.    \(v_0=2 \times 10^6\ \text{ms}^{-1}\)

c.i.    Point \(\text{Z.}\)

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
Show Worked Solution

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  
♦ Mean mark (a) 46%.

b.    \(v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}\)
 

c.i.    Point \(\text{Z.}\)
 

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
♦ Mean mark (c.i.) 40%.
♦♦♦ Mean mark (c.ii.) 15%.
COMMENT: Students needed to include what would happen to the electric force.

PHYSICS, M6 2021 VCE 5

The digram shows shows a stationary electron (e\(^{-}\)) in a uniform magnetic field between two parallel plates. The plates are separated by a distance of 6.0 × 10\(^{-3}\) m, and they are connected to a 200 V power supply and a switch. Initially, the plates are uncharged. Assume that gravitational effects on the electron are negligible.
 

  1. Explain why the magnetic field does not exert a force on the electron. Justify your answer with an appropriate formula.   (2 marks)

The switch is now closed.

  1. Determine the magnitude and the direction of any electric force now acting on the electron. Show your working.   (3 marks)

  1. Ravi and Mia discuss what they think will happen regarding the size and the direction of the magnetic force on the electron after the switch is closed.

    Ravi says that there will be a magnetic force of constant magnitude, but it will be continually changing direction.

    Mia says that there will be a constantly increasing magnetic force, but it will always be acting in the same direction.

    Evaluate these two statements, giving clear reasons for your answer.   (4 marks)

Show Answers Only

a.   Magnetic force does not exert force on electron:

  • Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).
  • Electron is stationary (\(v=0\))
  • The force on the electron = \(0\).

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

  • Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).

c.    Both statements contain a correct assertion and incorrect assertion.

  • Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.
  • Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.
  • This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 
  • Mia was incorrect to say that the direction of the magnetic force on the electron would not change.
  • Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.
  • This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.

Show Worked Solution

a.   Magnetic force does not exert force on electron:

  • Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).
  • Electron is stationary (\(v=0\))
  • The force on the electron = \(0\).
♦♦ Mean mark (a) 35%. 
COMMENT: Many students confused the forces of electric (not applicable here) and magnetic fields.

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

  • Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).

c.    Both statements contain a correct assertion and incorrect assertion.

  • Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.
  • Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.
  • This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 
  • Mia was incorrect to say that the direction of the magnetic force on the electron would not change.
  • Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.
  • This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.
♦♦♦ Mean mark (c) 18%.
COMMENT: Students need well planned responses to properly evaluate the physics principles required for 4 marks.

PHYSICS, M6 2022 VCE 3

A schematic diagram of a mass spectrometer that is used to deflect charged particles to determine their mass is shown in the diagram. Positive singly charged ions (with a charge of +1.602 × 10\(^{-19}\) C) are produced at the ion source. These are accelerated between an anode and a cathode. The potential difference between the anode and the cathode is 1500 V. The ions pass into a region of uniform magnetic field, \(B\), and are directed by the field into a semicircular path of diameter \(D\).
 

   

  1. Calculate the increase in the kinetic energy of each ion as it passes between the anode and the cathode. Give your answer in joules.  (2 marks)

Each ion has a mass of 4.80 × 10\(^{-27}\) kg.

  1. Show that each ion has a speed of 3.16 × 10\(^{5}\) m s\(^{-1}\) when it exits the cathode. Assume that the ion leaves the ion source with negligible speed. Show your working.  (2 marks)
  1. The region of uniform magnetic field, \(B\), in Figure 3 has a magnitude of 0.10 T.
  2. Calculate the diameter, \(D\), of the semicircular path followed by the ions within the magnetic field in Figure 3.  (3 marks)
Show Answers Only

a.    \(2.403 \times 10^{-16}\ \text{J}\)

b.     \(\Delta KE\) \(=\dfrac{1}{2}mv^2\)
  \(v\) \(=\sqrt{\dfrac{2 \Delta KE}{m}}\)
    \(=\sqrt{\dfrac{2 \times 2.403 \times 10^{-16}}{4.80 \times 10^{-27}}}\)
    \(=3.16 \times 10^5\ \text{ms}^{-1}\)

c.    \(0.19\ \text{m}\)

Show Worked Solution
a.    \(W=\Delta KE=qV=1.602 \times 10^{-19} \times 1500=2.403 \times 10^{-16}\ \text{J}\)
  
b.     \(\Delta KE\) \(=\dfrac{1}{2}mv^2\)
  \(v\) \(=\sqrt{\dfrac{2 \Delta KE}{m}}\)
    \(=\sqrt{\dfrac{2 \times 2.403 \times 10^{-16}}{4.80 \times 10^{-27}}}\)
    \(=3.16 \times 10^5\ \text{ms}^{-1}\)

  

c.     \(F_B\) \(=F_c\)
  \(qvB\) \(=\dfrac{mv^2}{r}\)
  \(r\) \(=\dfrac{mv}{qB}\)
    \(=\dfrac{4.80 \times 10^{-27} \times 3.16 \times 10^5}{1.602 \times 10^{-19} \times 0.10}\)
    \(=0.095\ \text{m}\)

 
\(\therefore\ D=0.19\ \text{m}\)

PHYSICS, M6 EQ-Bank 26

The diagram shows a stationary electron in a magnetic field. The magnetic field is surrounded by two parallel plates separated by a distance of `5.0 × 10^(-3) \ text {m}` and connected to a power supply and a switch.
 

The switch is initially open. At a later time the switch is closed.

Analyse the effects of the magnetic and electric fields on the acceleration of the electron both before and immediately after the switch is closed. In your answer, include calculation of the acceleration of the electron immediately after the switch is closed.   (5 marks)

Show Answers Only
  • Before the switch is closed, there is no electric field and so no electric force causing the electron to accelerate. As the electron is stationary, the magnetic field has no effect on it.
  • Once the switch is closed, the electric field causes the electron to accelerate down the page (towards the positive plate).
  • Calculating the acceleration of the electron immediately after the switch is closed:

`E=(V)/(d)=(100)/(0.005)=20\ 000\ text{V m}^(-1)`

`F=Eq=20\ 000 xx1.602 xx10^(-19)=3.2 xx10^(-15)  text{N}`

`a=(F)/(m)=(3.2 xx10^(-15))/(9.109 xx10^(-31))=3.5 xx10^(15)  text{m s}^(-2)  text{down page.}`

  • Now that the electron is moving, the magnetic field exerts a force on it towards the right (using the right hand palm rule).
  • The force and acceleration on the electron will increase due to its increasing velocity.
Show Worked Solution
  • Before the switch is closed, there is no electric field and so no electric force causing the electron to accelerate. As the electron is stationary, the magnetic field has no effect on it.
  • Once the switch is closed, the electric field causes the electron to accelerate down the page (towards the positive plate).
  • Calculating the acceleration of the electron immediately after the switch is closed:

`E=(V)/(d)=(100)/(0.005)=20\ 000\ text{V m}^(-1)`

`F=Eq=20\ 000 xx1.602 xx10^(-19)=3.2 xx10^(-15)  text{N}`

`a=(F)/(m)=(3.2 xx10^(-15))/(9.109 xx10^(-31))=3.5 xx10^(15)  text{m s}^(-2)  text{down page.}`

  • Now that the electron is moving, the magnetic field exerts a force on it towards the right (using the right hand palm rule).
  • The force and acceleration on the electron will increase due to its increasing velocity.

PHYSICS, M6 EQ-Bank 8 MC

A positively-charged ion travelling at 250 ms ¯1 is fired between two parallel charged plates, \(M\) and \(N\). There is also a magnetic field present in the region between the two plates. The direction of the magnetic field is into the page as shown. The ion is travelling perpendicular to both the electric and the magnetic fields.
 

The electric field between the plates has a magnitude of 200 V m ¯1. The magnetic field is adjusted so that the ion passes through undeflected.

What is the magnitude of the adjusted magnetic field, and the polarity of the \(M\) terminal relative to the \(N\) terminal?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Magnitude of magnetic field }& \textit{Polarity of M relative to N} \\
\text{(teslas)}\rule[-1ex]{0pt}{0pt}& \text{} \\
\hline
\rule{0pt}{2.5ex}0.8\rule[-1ex]{0pt}{0pt}&\text{positive}\\
\hline
\rule{0pt}{2.5ex}0.8\rule[-1ex]{0pt}{0pt}& \text{negative}\\
\hline
\rule{0pt}{2.5ex}1.25\rule[-1ex]{0pt}{0pt}& \text{positive} \\
\hline
\rule{0pt}{2.5ex}1.25\rule[-1ex]{0pt}{0pt}& \text{negative} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • The ion passes through undeflected, so the magnitude of the force it experiences due to the magnetic field is equal to the magnitude of the force it experiences due to the electric field. 
  • As the ion travels perpendicular to the magnetic field:
\(F\) \(=qvB\)  
\(E q\) \(=qvB\)  
\(B\) \(=\dfrac{E}{v}\)  
  \(=\dfrac{V}{d} \times \dfrac{1}{v}=\dfrac{200}{250}=0.8 \ \text{T}\)  
     
  •  Using the right hand palm rule, the magnetic field exerts a force up the page on the positive ion. The electric field must therefore exert a force down the page on the ion.
  • \(M\) must be positive relative to \(N\).

\(\Rightarrow A\)

PHYSICS, M6 2016 HSC 3 MC

A region of space contains a constant magnetic field and a constant electric field.

How will these fields affect an electron that is stationary in this region?

  1. Both fields will exert a force.
  2. Neither field will exert a force.
  3. Only the electric field will exert a force.
  4. Only the magnetic field will exert a force.
Show Answers Only

`C`

Show Worked Solution
  • Magnetic fields only exert a force on moving charged particles while electric fields exert a force on all charged particles moving or stationary.

`=>C`


♦♦ Mean mark 35%.

PHYSICS, M6 2017 HSC 30

In a thought experiment, a proton is travelling at a constant velocity in a vacuum with no field present. An electric field and a magnetic field are then turned on at the same time.

The fields are uniform in magnitude and direction and can be considered to extend infinitely. The velocity of the proton at the instant the fields were turned on is perpendicular to the fields.
 

Analyse the motion of the proton after the fields have been turned on.   (4 marks)

Show Answers Only
  • Using the right hand palm rule, the magnetic field exerts a force out of the page, perpendicular to the velocity of the proton.
  • So, the magnetic field causes the proton to undergo circular motion, initially moving out of the page and continuing in an anti-clockwise direction as viewed from the right.
  • The electric field exerts a constant force to the left on the proton.
  • This causes the proton to accelerate towards the left. The resultant motion will be a helix (vector sum of motion) that extends to the left, with the distance between adjacent spirals increasing.
  • The helix will decrease in radius as the proton loses kinetic energy and hence speed as it radiates electromagnetic waves. This occurs because the proton is an accelerating charge.

Show Worked Solution
  • Using the right hand palm rule, the magnetic field exerts a force out of the page, perpendicular to the velocity of the proton.
  • So, the magnetic field causes the proton to undergo circular motion, initially moving out of the page and continuing in an anti-clockwise direction as viewed from the right.
  • The electric field exerts a constant force to the left on the proton.
  • This causes the proton to accelerate towards the left. The resultant motion will be a helix (vector sum of motion) that extends to the left, with the distance between adjacent spirals increasing.
  • The helix will decrease in radius as the proton loses kinetic energy and hence speed as it radiates electromagnetic waves. This occurs because the proton is an accelerating charge.

♦ Mean mark 51%.

PHYSICS, M6 2018 HSC 12 MC

The diagram shows electrons travelling in a vacuum at  `2 × 10^6 \ text{m s}^(-1)`  between two charged metal plates  `1 × 10^-3\ text{m}`  apart.
 

A magnetic field is to be applied to make the electrons continue to travel in a straight line.

What is the magnitude and direction of the magnetic field that is to be applied?

  1. `5 × 10^-1 \ text {T}`  into the page
  2. `5  × 10^-1 \ text {T}`  out of the page
  3. `1 × 10^6 \ text {T}`  into the page
  4. `1 × 10^6 \ text {T}`  out of the page
Show Answers Only

`A`

Show Worked Solution
  • There will be an upwards force (towards the positive plate) on the electrons due to the electric field.
  • Using the right hand palm rule, the magnetic field must be into the page to produce a downwards force and balance the upwards force.
  • Since the magnitudes of electric and magnetic force are equal:
`qE` `=qvB`  
`B` `=(E)/(v)=((1000)/(1xx10^(-3)))/(2xx10^(6))=5xx10^(-1)\ text{T}`  

 
`=>A`


♦ Mean mark 48%.

PHYSICS M6 2022 HSC 12 MC

The diagram shows a region in which there are uniform electric and magnetic fields. A positively charged particle moves in the region at constant velocity.
 


 

What is the direction of the particle's velocity?

  1. Up the page
  2. Down the page
  3. To the left
  4. To the right
Show Answers Only

`D`

Show Worked Solution
  • For the particle to have a constant velocity, the net force acting on it must be zero.
  • The electric field exerts a downwards force on the particle.
  • The magnetic field must exert an upwards force on the particle.
  • Using the right hand palm rule, the direction of the particle’s velocity is to the right.

`=>D`


♦ Mean mark 52%.

PHYSICS, M6 2020 HSC 10 MC

An electron travelling in a straight line with an initial velocity, `u`, enters a region between two charged plates in which there is an electric field causing it to travel along the path as shown.
 

A magnetic field is then applied causing a second electron with the same initial velocity to pass through undeflected.

Which row of the table shows the directions of the electric and magnetic fields when the second electron enters the region between the plates?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \textit{Electric Field}\quad \rule[-1ex]{0pt}{0pt}& \quad \textit{Magnetic Field} \quad \\
\hline
\rule{0pt}{2.5ex}\text{Towards top of page}\rule[-1ex]{0pt}{0pt}&\text{Into page}\\
\hline
\rule{0pt}{2.5ex}\text{Towards top of page}\rule[-1ex]{0pt}{0pt}& \text{Out of page}\\
\hline
\rule{0pt}{2.5ex}\text{Towards bottom of page}\rule[-1ex]{0pt}{0pt}& \text{Into page} \\
\hline
\rule{0pt}{2.5ex}\text{Towards bottom of page}\rule[-1ex]{0pt}{0pt}& \text{Out of page} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • The bottom plate is positively charged as it attracts the electron.
  • Electric field is towards the top of the page.
  • Using the right hand palm rule, the magnetic field must be out of the page to produce an upwards force on the electron.

\(\Rightarrow B\)


♦ Mean mark 50%.