SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

PHYSICS, M6 2019 VCE 1

A particle of mass \(m\) and charge \(q\) travelling at velocity \(v\) enters a uniform magnetic field \(\text{B}\), as shown in Figure 1.
 

  1. Is the charge \(q\) positive or negative? Give a reason for your answer.   (1 mark)
  1. Explain why the path of the particle is an arc of a circle while the particle is in the magnetic field.   (2 marks)

Show Answers Only

a.    Using the right-hand rule:

→ Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).

→ As the force on the charge is down the page (back of the hand), the charge must be negative.
 

b.    For circular motion to occur:

→ The force must be at right-angles to the velocity of the particle.

→ The force must be of constant magnitude \((F=qvB)\).

→ In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.

→ The particle will therefore follow the arc of a circle while in the magnetic field.

Show Worked Solution

a.    Using the right-hand rule:

→ Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).

→ As the force on the charge is down the page (back of the hand), the charge must be negative.

♦ Mean mark (a) 44%.

b.    For circular motion to occur:

→ The force must be at right-angles to the velocity of the particle.

→ The force must be of constant magnitude \((F=qvB)\).

→ In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.

→ The particle will therefore follow the arc of a circle while in the magnetic field.

♦♦♦ Mean mark (b) 27%.

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 2020 VCE 3-4 MC

A positron with a velocity of 1.4 × 10\(^6\) m s\(^{-1}\) is injected into a uniform magnetic field of 4.0 × 10\(^{-2}\) T, directed into the page, as shown in the diagram below.

It moves in a vacuum in a semicircle of radius \(r\). The mass of the positron is 9.1 × 10\(^{-31}\) kg and the charge on the positron is 1.6 × 10\(^{-19}\) C. Ignore relativistic effects.
 


 

Question 3

Which one of the following best gives the speed of the positron as it exits the magnetic field?

  1. 0 m s\(^{-1}\)
  2. much less than 1.4 × 10\(^6\) m s\(^{-1}\)
  3. 1.4 × 10\(^6\) m s\(^{-1}\)
  4. greater than 1.4 × 10\(^6\) m s\(^{-1}\)

 
Question 4

The speed of the positron is changed to 7.0 × 10\(^5\) m s\(^{-1}\).

Which one of the following best gives the value of the radius \(r\) for this speed?

  1. \(\dfrac{r}{4}\)
  2. \(\dfrac{r}{2}\)
  3. \(r\)
  4. \(2 r\)
Show Answers Only

\(\text{Question 3:}\ C\)

\(\text{Question 4:}\ B\)

Show Worked Solution

\(\text{Question 3}\)

→ As the direction of the force is perpendicular to the velocity, the velocity will not change.

→ The velocity will be \(1.4 \times 10^6\ \text{ms}^{-1}\)

\(\Rightarrow C\)
 

\(\text{Question 4}\)

\(F_B\) \(=F_c\)  
\(qvB\) \(=\dfrac{mv^2}{r}\)  
\(r\) \(=\dfrac{mv}{qB}\)  

 
→ Therefore, \(r \propto v\).

→ If the velocity is halved, the radius will also be halved.

\(\Rightarrow B\)

PHYSICS, M6 2021 VCE 5

Figure 5 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.    → 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.    → 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 explain 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 Figure 3. 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.

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 2023 VCE 1 MC

One type of loudspeaker consists of a current-carrying coil within a radial magnetic field, as shown in the diagram below. \(X\) and \(Y\) are magnetic poles, and the direction of the current, \(I\), in the coil is clockwise as shown.
 


 

The force, \(F\), acting on the current-carrying coil is directed into the page.

Which one of the following statements correctly identifies the magnetic polarities of \(X\) and \(Y\)?

  1. \(X\) is a north pole and \(Y\) is a south pole.
  2. \(X\) is a south pole and \(Y\) is a north pole.
  3. Both \(X\) and \(Y\) are north poles.
  4. Both \(X\) and \(Y\) are south poles.
Show Answers Only

\(A\)

Show Worked Solution

→ Using the right hand rule, the palm faces into the page (direction of the force) and thumb points down the page (direction of the current).

→ The direction of the magnetic field is from \(X\) and \(Y\) (direction of fingers).

→ \(X\) is a north pole and \(Y\) is a south pole.

\(\Rightarrow A\)

PHYSICS, M6 2023 HSC 24

An electron is travelling at 3.0 \(\times\) 10\(^{6}\) m s\(^{-1}\) in the path shown.
 

Calculate the magnetic field required to keep the electron in the path.  (3 marks)

Show Answers Only

\(1.7 \times 10^{-6}\) \(\text{T}\)

Show Worked Solution

→ The force from the magnetic field on the electron provides the centripetal acceleration for it to travel in uniform circular motion.

\( F_B\) \(=F_c\)  
\(qvB\) \(=\dfrac{mv^2}{r}\)  
\(B\) \(=\dfrac{mv}{qr}\)  
  \(=\dfrac{9.109 \times 10^{-31} \times 3.0 \times 10^6}{1.602 \times 10^{-19} \times 10}\)  
  \(=1.7 \times 10^{-6}\) \(\text{T}\)  

 
→ Magnetic field strength required = \(1.7 \times 10^{-6}\) \(\text{T}\).

Mean mark 57%.

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.

Show Answers Only

“

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 the 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?
 

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`  
`Eq` `=qvB`  
`B` `=(E)/(v)`  
  `=(V)/(d)((1)/(v))`    `(E=(V)/(d))`
  `=(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`.

`=>A`

PHYSICS, M6 2015 HSC 8 MC

In which of the following situations does the magnetic field exert the greatest force on the proton ( ), given that all of the fields are of equal magnitude?

 

 

Show Answers Only

`C`

Show Worked Solution

→ The force on each proton is given by  `F=qvB sin theta.`

→ As `q` and `B` are the same in all four situations, the maximum force will occur when  `v sin theta`  is greatest.

A:  `vsintheta=0`

B:  `vsintheta=60sin 0°=0`

C:  `vsintheta=40sin 90°=40`

D:  `vsintheta=50sin 45°~~35`

`=>C`

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 2017 HSC 18 MC

A particle of mass `m` and charge `q` travelling at velocity `v` enters a magnetic field of magnitude `B` and follows the path shown.
 

A second particle enters a magnetic field of magnitude `2B` with a velocity of `(1)/2 v` and follows an identical path.

What is the mass and charge of the second particle?
 

Show Answers Only

`C`

Show Worked Solution

→ The centripetal force acting on the charge is given by the force it experiences due to the magnetic field:

`F_(c)` `=F_(b)`  
`(mv^2)/(r)` `=qvB`  
`r` `=(mv)/(qB)`  

 
Given `B=2B`  and  `v=(1)/(2)v`:

`r` `=(m(1)/(2)v)/(q(2B))`  
`r` `=(mv)/(4qB)`  

 
→ In order for the radius to remain the same, `(m)/(q)` must be four times what it was originally.

`=>C`


♦♦ Mean mark 35%.

PHYSICS, M6 2018 HSC 13 MC

An electron moves in a circular path with radius `r` in a magnetic field as shown.
 

If the speed of the electron is increased, which row of the table correctly shows the effects of this change?
 

Show Answers Only

`B`

Show Worked Solution

→ Using the formula  `F=qvB`, increasing the speed of the electron increases the force acting on it.

→ The centripetal force acting on the electron is given by the force it experiences due to the magnetic field:

`(mv^2)/(r)=qvB\ \ =>\ \ r=(mv)/(qB)\ \ =>\ \ r prop v`

→ Increasing the speed of the electron increases its radius.

`=>B`


Mean mark 53%.

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 2019 HSC 33

A proton and an alpha particle are fired into a uniform magnetic field with the same speed from opposite sides as shown. Their trajectories are initially perpendicular to the field.
 


  

Explain ONE similarity and ONE difference in their trajectories as they move in the magnetic field.   (4 marks)

Show Answers Only

Similarity:

→ Both particles experience a constant force, given by  `F=qvB`, perpendicular to their velocity and the magnetic field lines and will undergo circular motion.

Difference:

→ The centripetal force acting on both particles is given by the force they experience due to the magnetic field as follows:

`(mv^2)/(r)=qvB\ \ =>\ \ r=(mv)/(qB)`

→ The alpha particle has four times the mass and two times the charge of the proton

→ Therefore, the radius of its trajectory will be twice that of the protons.

Show Worked Solution

Similarity:

→ Both particles experience a constant force, given by  `F=qvB`, perpendicular to their velocity and the magnetic field lines and will undergo circular motion.

Difference:

→ The centripetal force acting on both particles is given by the force they experience due to the magnetic field as follows:

`(mv^2)/(r)=qvB\ \ =>\ \ r=(mv)/(qB)`

→ The alpha particle has four times the mass and two times the charge of the proton

→ Therefore, the radius of its trajectory will be twice that of the protons.

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 19 MC

A conductor `P Q` is in a uniform magnetic field. The conductor rotates around the end `P` at a constant angular velocity.
 

Which graph shows the induced emf between `P` and `Q` as the conductor completes one revolution from the position shown?
 

 

Show Answers Only

`C`

Show Worked Solution

At the starting position shown, electrons in the rod are moving to the right, parallel to the magnetic field lines. So, there is no force acting on the.

→ EMF of zero.

After a quarter of a rotation, electrons in the rod are moving up the page. Using the right hand palm rule, they experience a force out of the page. This will not induce an EMF between `P` and `Q.`

→ the graph will show an EMF of zero at both `t=0` and after a quarter of a rotation.

`=>C`


♦♦ Mean mark 10%.

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?
 

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.

`=>B`


♦ Mean mark 50%.