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PHYSICS, M6 2024 HSC 33

A magnet is swinging as a pendulum. Close below it is an aluminium (non-ferromagnetic) can. The can is free to spin around a fixed axis as shown.
 

Analyse the motion and energy transformations of both the can and the magnet.   (7 marks)

Show Answers Only

  • When the magnet swings down from its high position toward the can, its gravitational potential energy transforms into kinetic energy.
  • As the magnet moves, it creates changing magnetic flux through the aluminium can. This flux change is strongest when there’s the fastest relative motion between the magnet and can.
  • The induced emf is described in the equation  \({\large{\varepsilon}} = -N \dfrac{\Delta \phi}{\Delta t} \).
  • This emf creates eddy currents in the can, which produce both heat and a magnetic field. Following Lenz’s law, this magnetic field opposes the magnet’s motion.
  • The magnetic fields from both the magnet and the eddy currents interact, causing the can to initially rotate clockwise.
  • Eventually, this interaction dampens the magnet’s swing. The magnetic interaction between the eddy currents and the magnet causes the can to rotate back and forth with decreasing amplitude, as the system’s energy gradually converts to heat.

Show Worked Solution

  • When the magnet swings down from its high position toward the can, its gravitational potential energy transforms into kinetic energy.
  • As the magnet moves, it creates changing magnetic flux through the aluminium can. This flux change is strongest when there’s the fastest relative motion between the magnet and can.
  • The induced emf is described in the equation  \({\large{\varepsilon}} = -N \dfrac{\Delta \phi}{\Delta t} \).
  • This emf creates eddy currents in the can, which produce both heat and a magnetic field. Following Lenz’s law, this magnetic field opposes the magnet’s motion.
  • The magnetic fields from both the magnet and the eddy currents interact, causing the can to initially rotate clockwise.
  • Eventually, this interaction dampens the magnet’s swing. The magnetic interaction between the eddy currents and the magnet causes the can to rotate back and forth with decreasing amplitude, as the system’s energy gradually converts to heat.
♦♦ Mean mark 47%.

PHYSICS, M6 2024 HSC 18 MC

The diagram shows a magnet moving towards a coil \(X\).
 

This action causes a current to be induced in the coil.

Which situation will induce a current in coil \(X\) that is in the same direction as the current induced by the movement of the magnet?
 

Show Answers Only

\(D\)

Show Worked Solution
  • As the north pole of the magnet moves towards coil \(X\), the magnetic flux through \(X\) is increased. By Lenz’s law, a current will be generated in the coil that produces a magnetic field to oppose the increase the flux. Using Lenz’s law, the current in \(X\) will generate a north pole to the left to repel the approaching magnet. The induced current will run up the coils as viewed from the front. 

  • The current will be induced in the same direction in \(D\). Using the right hand grip rule, the current though the white coil will produce magnetic field lines gong to the left. As the current is decreasing, the change of flux through \(X\) is decreasing. Therefore by Lenz’s law, the current induced in \(X\) will strengthen the magnetic field to oppose the decreasing magnetic flux. 
  • Therefore, the magnetic field lines produced by the current in \(X\) will also be to the left, and using the right hand grip rule, the current through \(X\) will run up the coils (front view).
  • In all the other options, the current will run down the coils (front view).

\(\Rightarrow D\)

♦♦♦ Mean mark 28%.

PHYSICS, M6 2020 VCE 6

Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.
 

  1. Will the magnetic flux through the coil increase, decrease or stay the same as the students change the shape of the coil?   (1 mark)
  1. Explain, using physics principles, why the ammeter registered a current in the coil and determine the direction of the induced current.   (3 marks)
  1. The students then push each side of the coil together, as shown in Figure 6a, so that the coil returns to its original circular shape, as shown in Figure 6b, and then changes to the shape shown in Figure 6c. 
     

  1. Describe the direction of any induced currents in the coil during these changes. Give your reasoning.   (2 marks)

Show Answers Only

a.    Decrease 

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
Show Worked Solution

a.    Magnetic flux will decrease.

  • The magnetic flux is proportional to how many field lines are passing through a given area.
  • As the number of lines passing through the coil of wire decreases, the magnetic flux will also decrease.

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.
♦ Mean mark (b) 42%.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
♦♦♦ Mean mark (c) 28%.
COMMENT: Lenz’s law was poorly understood in this question.

PHYSICS, M6 2023 VCE 5

Figure 1 shows a single square loop of conducting wire placed just outside a constant uniform magnetic field, \(B\). The length of each side of the loop is 0.040 m. The magnetic field has a magnitude of 0.30 T and is directed out of the page.

Over a time period of 0.50 s, the loop is moved at a constant speed, \(v\), from completely outside the magnetic field, Figure 1, to completely inside the magnetic field, Figure 2.
 

  1. Calculate the average EMF produced in the loop as it moves from the position just outside the region of the field, Figure 1, to the position completely within the area of the magnetic field, Figure 2.
  2. Show your working.   (2 marks)
  1. On the small square loop in Figure 3, show the direction of the induced current as the loop moves into the area of the magnetic field.   (1 mark)
     

Show Answers Only

a.    \(9.6 \times 10^{-4}\ \text{V}\)

b.    
       

Show Worked Solution
a.     \(\varepsilon\) \(=\dfrac{\Delta \phi}{\Delta t}\)
    \(=\dfrac{\Delta B A}{\Delta t}\)
    \(=\dfrac{0.3 \times 0.04^2}{0.5}\)
    \(=9.6 \times 10^{-4}\ \text{V}\)

 

b.  
         

♦ Mean mark (b) 45%.

PHYSICS, M6 2023 HSC 30

The diagram shows apparatus that is used to investigate the interaction between the magnetic field produced by a coil and two copper rings \(X\) and \(Y\), when each is placed at position \(P\), as shown.
 

 

Ring \(X\) is a complete circular ring, and a small gap has been cut in ring \(Y\). Each of the rings has a cross-sectional area of  4 × 10\(^{-4}\) m².

The power supply connected to the coil produces an increasing current through the coil in the direction shown, when the switch is turned on. This produces a magnetic field at \(P\) that varies as shown in the graph.
 

  1. In the first part of the investigation, ring \(X\) is held near the end of the electromagnet at position \(P\).
  2. Account for the force acting on the ring from 0 to 0.05 seconds after the power supply is turned on.   (4 marks)
  1. i.  In the second part of the investigation, ring \(Y\) is placed at \(P\), and the power supply is turned on. 
  2.     Explain the behaviour of the ring.   (2 marks)
  1. ii. Calculate the maximum induced emf in ring \(Y\).   (2 marks)

Show Answers Only

a.   Between 0 – 0.03 seconds:

  • The magnetic field strength at point \(P\) is increasing at a constant rate.
  • Thus, ring \(X\) experiences a change in flux, which causes an EMF to be induced across the ring (Faraday’s Law) 
  • This induced EMF causes a current to flow through ring \(X\) that will interact in the external magnetic field to reduce the change in flux that created it (Lenz’s Law).
  • Thus, a clockwise current will run through the ring, creating a north pole towards the solenoid, causing the ring to have a repulsive force away from the solenoid acting on it.

Between 0.03 – 0.05 seconds: 

  • There is no change in flux through the ring due to there being a constant magnetic field.
  • Therefore, there is no induced EMF or induced current.
  • Therefore, there is no force acting on the ring.

b.i.  Ring \(Y\) behaviour when placed at \(P\):

  • Ring \(Y\) will experience the same change in flux and hence the same induced EMF as ring \(X\) within 0 – 0.03 seconds.
  • However, due to the gap in ring \(Y\) (i.e. there is no closed circuit), no induced current will be able to flow through the ring.
  • Hence, there is no force exerted on ring \(Y\).

ii.  \(\varepsilon = 8 \times 10^{-5}\ \text{V}\)

Show Worked Solution

a.   Between 0 – 0.03 seconds:

  • The magnetic field strength at point \(P\) is increasing at a constant rate.
  • Thus, ring \(X\) experiences a change in flux, which causes an EMF to be induced across the ring (Faraday’s Law) 
  • This induced EMF causes a current to flow through ring \(X\) that will interact in the external magnetic field to reduce the change in flux that created it (Lenz’s Law).
  • Thus, a clockwise current will run through the ring, creating a north pole towards the solenoid, causing the ring to have a repulsive force away from the solenoid acting on it.

Between 0.03 – 0.05 seconds: 

  • There is no change in flux through the ring due to there being a constant magnetic field.
  • Therefore, there is no induced EMF or induced current.
  • Therefore, there is no force acting on the ring.
♦ Mean mark (a) 44%.

b.i.  Ring \(Y\) behaviour when placed at \(P\):

  • Ring \(Y\) will experience the same change in flux and hence the same induced EMF as ring \(X\) within 0 – 0.03 seconds.
  • However, due to the gap in ring \(Y\) (i.e. there is no closed circuit), no induced current will be able to flow through the ring.
  • Hence, there is no force exerted on ring \(Y\).
b.ii. \(\varepsilon\) \(=N\dfrac{\Delta \phi}{\Delta t}\)
    \(=N \dfrac{A \Delta B}{\Delta t}\)
    \(= 1 \times \dfrac{4 \times 10^{-4} \times (6 \times 10^{-3}-0)}{0.03-0}\)
    \(=8 \times 10^{-5}\ \text{V}\)
♦♦ Mean mark (b)(i) 38%.
♦ Mean mark (b)(ii) 49%.

PHYSICS, M6 2015 HSC 28

A copper plate is attached to a lightweight trolley. The trolley moves at an initial velocity, `v`, towards a strong magnet fixed to a support.
 

The dashed line on the graph shows the velocity of the trolley when the magnet is not present.

On the axes, sketch the graph of the velocity of the trolley as it travels from `A` to `D` under the magnet, and justify your graph.   (5 marks)
 

Show Answers Only

  • As the trolley passes under the magnet at B, its kinetic energy is transformed into heat energy in the copper plate.
  • This causes the trolley to slow down.
  • This result is due to the change in flux produced by the movement of the copper plate through the field of the strong magnet.
  • The change in flux induces currents in the copper plate (Faraday’s law) that produce a magnetic field that opposes the changing flux (Lenz’s law). The resulting force decelerates the trolley.
Show Worked Solution

  • As the trolley passes under the magnet at B, its kinetic energy is transformed into heat energy in the copper plate.
  • This causes the trolley to slow down.
  • This result is due to the change in flux produced by the movement of the copper plate through the field of the strong magnet.
  • The change in flux induces currents in the copper plate (Faraday’s law) that produce a magnetic field that opposes the changing flux (Lenz’s law). The resulting force decelerates the trolley.

PHYSICS, M6 2015 HSC 15 MC

A circular loop of wire is stationary in a magnetic field. The sides are then pushed together to change the shape, as shown in the diagram.
 

As the loop is compressed, a current is induced.

Which row of the table shows the direction of the current and explains why it is induced?

\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|l|}
\hline
\rule{0pt}{2.5ex}\textit{Current Direction}\rule[-1ex]{0pt}{0pt}& \textit{Why the current is induced} \\
\hline
\rule{0pt}{2.5ex}\text{Clockwise}\rule[-1ex]{0pt}{0pt}&\text{Change in magnetic flux}\\
\hline
\rule{0pt}{2.5ex}\text{Anticlockwise}\rule[-1ex]{0pt}{0pt}& \text{Change in magnetic flux}\\
\hline
\rule{0pt}{2.5ex}\text{Clockwise}\rule[-1ex]{0pt}{0pt}& \text{Change in magnetic flux density} \\
\hline
\rule{0pt}{2.5ex}\text{Anticlockwise}\rule[-1ex]{0pt}{0pt}& \text{Change in magnetic flux density} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • Compressing the loop decreases the area of the loop.
  • There is a change (reduction) in magnetic flux passing through the loop.
  • A current is generated in the loop in order to mitigate this change by producing a magnetic field directed into the page.
  • Using the right hand grip rule, the current direction is clockwise.

\(\Rightarrow A\)


♦♦ Mean mark 38%.

PHYSICS, M6 2016 HSC 30

The following makeshift device was made to provide lighting for a stranded astronaut on Mars.

The mass of Mars is `6.39 ×10^(23) \ text {kg}`.
 

The 2 kg mass falls, turning the DC generator, which supplies energy to the light bulb. The mass falls from a point that is 3 376 204 m from the centre of Mars.

  1. Calculate the maximum possible energy released by the light bulb as the mass falls through a distance of one metre.   (3 marks)
  1. Explain the difference in the behaviour of the falling mass when the switch is open.   (3 marks)
Show Answers Only

a.   7.48 J

b.   When switch is opened:

  • There is no induced current opposing the downwards motion of the mass (Lenz’s Law).
  • Hence, the mass will fall more quickly.
Show Worked Solution

a.   `DeltaE=U_(f)-U_(i)`

`=((-Gm_(1)m_(2))/(r_(f)))-((-Gm_(1)m_(2))/(r_(i)))`

`=(-6.67 xx10^(-11)xx6.39 xx10^(23)xx2)/(3\ 376\ 203)-((-6.67 xx10^(-11)xx6.39 xx10^(23)xx2))/(3\ 376\ 204)`

`=-7.48\ text{J}`
 

  • 7.48 J is lost by the falling mass.
  • The light bulb released 7.48 J of energy.

♦ Mean mark (a) 52%.

b.   When switch is opened:

  • There is no induced current opposing the downwards motion of the mass (Lenz’s Law).
  • Hence, the mass will fall more quickly.

♦♦♦ Mean mark (b) 27%.

PHYSICS, M6 2016 HSC 22a

When an alternating current is passed through coil `A`, a voltage is observed on the oscilloscope connected to coil `B`.
 
     
 

How could a bar magnet be used, instead of coil A, to produce a similar pattern on the oscilloscope?   (2 marks)

Show Answers Only

Answers could include one of the following:

  • Move the bar magnet repeatedly towards and away from the end of coil `B`.
  • Rotate the bar magnet in the plane of the page.
  • Any other motion which induces an oscillating voltage in coil `B`.
Show Worked Solution

Answers could include one of the following:

  • Move the bar magnet repeatedly towards and away from the end of coil `B`.
  • Rotate the bar magnet in the plane of the page.
  • Any other motion which induces an oscillating voltage in coil `B`.

PHYSICS, M6 2016 HSC 22b

A strong magnet is at rest a few centimetres above a solid metal disc made of a non-magnetic metal. The magnet is then dropped.   
 
   
 

The velocity of the magnet is shown in this graph.
 
   
 

Account for the shape of the graph.   (4 marks)

Show Answers Only
  • Initially, the magnet falls only under the influence of gravity accelerating downwards at a rate of 9.8 m s –1.
  • This corresponds to the downwards line on the graph with a gradient of – 9.8.
  • As the magnet approaches the disc, eddy currents are induced in the disc due to the changing flux (Faraday’s Law).
  • The magnetic field formed by these eddy currents opposes the motion of the disc (Lenz’s Law).
  • This slows the descent of the magnet, corresponding to the upwards line on the graph.
  • Eventually, the magnet comes to rest on the disc, shown when `v=0` on the graph.
Show Worked Solution
  • Initially, the magnet falls only under the influence of gravity accelerating downwards at a rate of 9.8 m s –1.
  • This corresponds to the downwards line on the graph with a gradient of – 9.8.
  • As the magnet approaches the disc, eddy currents are induced in the disc due to the changing flux (Faraday’s Law).
  • The magnetic field formed by these eddy currents opposes the motion of the disc (Lenz’s Law).
  • This slows the descent of the magnet, corresponding to the upwards line on the graph.
  • Eventually, the magnet comes to rest on the disc, shown when `v=0` on the graph.

PHYSICS, M6 2016 HSC 8 MC

Which movement of the magnet(s) will produce the greatest deflection of the galvanometer?
 

 

 

Show Answers Only

`D`

Show Worked Solution
  • A larger induced current will produce greater deflection of the galvanometer.
  • The induced current is proportional to the total rate of change of magnetic flux through the coil.
  • The faster the movement of the magnet, the greater the rate of change of flux.

`=>D`

PHYSICS, M6 2016 HSC 7 MC

A magnet passes through a copper tube at constant velocity along the path shown.

A current is induced in the tube by the motion of the magnet.
 

Which row of the table correctly describes the forces acting between the tube and the magnet at points \(P\) and \(Q\) ? 
 

\begin{align*}
\begin{array}{l}
& \\
\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
\quad \textit{Force at P} \quad & \quad \textit{Force at Q} \quad \\
\hline
\rule{0pt}{2.5ex}\text{Attraction}\rule[-1ex]{0pt}{0pt}&\text{Repulsion}\\
\hline
\rule{0pt}{2.5ex}\text{Repulsion}\rule[-1ex]{0pt}{0pt}& \text{Attraction}\\
\hline
\rule{0pt}{2.5ex}\text{Attraction}\rule[-1ex]{0pt}{0pt}& \text{Attraction} \\
\hline
\rule{0pt}{2.5ex}\text{Repulsion}\rule[-1ex]{0pt}{0pt}& \text{Repulsion} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • Lenz’s Law states that the force due to the induced current in the tube will oppose the motion of the magnet.
  • At \(P\) a force of repulsion will act to oppose the motion of the magnet towards the tube.
  • At \(Q\) a force of attraction will act to oppose the motion of the magnet away from the tube.

\(\Rightarrow B\)

PHYSICS, M6 2017 HSC 19 MC

Earth's magnetic field is shown in the following diagram.
 

Two students standing a few metres apart on the equator at points `X` and `Y`, where Earth's magnetic field is parallel to the ground, hold a loop of copper wire between them. Part of the loop is rotated like a skipping rope as shown, while the other part remains motionless on the ground.
 

At what point during the rotation of the wire does the maximum current flow in a direction from `P` to `Q` through the moving part of the wire?

  1. `A`
  2. `B`
  3. `C`
  4. `D`
Show Answers Only

`C`

Show Worked Solution
  • The magnetic field through the loop is going into the page.
  • Using the right hand palm rule, the maximum current flow from `P` to `Q` produces an upwards force on the wire.
  • By Lenz’s Law, this occurs when the wire is moving downwards.

`=>C`


♦♦ Mean mark 28%.

PHYSICS, M6 2018 HSC 22

  1. A drill spins a magnet above a non-magnetic metal disc which is free to rotate.
     
       
     
    Explain the effect of the rotating magnet on the disc.   (3 marks)
  1. The diagram shows a magnet attached to an electric drill so that it can be rotated between two coils connected to a voltmeter.
     
       
     
    The drill starts from rest and gradually speeds up, reaching its full speed after three revolutions.
     
    Sketch a graph showing the induced emf across the coils during the time that it takes the magnet to reach its full speed.   (3 marks)
     
Show Answers Only

a.    Effect of rotating magnet:

  • The rotating magnet causes a changing magnetic flux in the metal disc.
  • This induces eddy currents (Faraday’s Law). By Lenz’s Law, these produce a magnetic field which opposes the change in flux by minimising the relative motion between the disc and the magnet.
  • This will have the effect of the disc rotating in the same direction as the magnet.

b.   

         

Show Worked Solution

a.    Effect of rotating magnet:

  • The rotating magnet causes a changing magnetic flux in the metal disc.
  • This induces eddy currents (Faraday’s Law). By Lenz’s Law, these produce a magnetic field which opposes the change in flux by minimising the relative motion between the disc and the magnet.
  • This will have the effect of the disc rotating in the same direction as the magnet.

b.   

         


♦ Mean mark (b) 50%.

PHYSICS, M6 2018 HSC 18 MC

An experiment is set up as shown.
 

When the switch is closed, the reading on the spring balance changes immediately, then returns to the initial reading.

Which row of the table correctly shows the direction of the current through the straight conductor \(XY\) and the direction in which the pointer on the spring balance initially moves?

\begin{align*}
\begin{array}{l}
\textit{}& \textit{} \\
\textit{}& \textit{} \\
\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
\textit{Direction of current through}& \textit{Direction in which the} \\
\textit{the straight conductor}& \textit{pointer initially moves} \\
\hline
\rule{0pt}{2.5ex}\text{From \(X\) to \(Y\)}\rule[-1ex]{0pt}{0pt}&\text{Down}\\
\hline
\rule{0pt}{2.5ex}\text{From \(X\) to \(Y\)}\rule[-1ex]{0pt}{0pt}& \text{Up}\\
\hline
\rule{0pt}{2.5ex}\text{From \(Y\) to \(X\)}\rule[-1ex]{0pt}{0pt}& \text{Down} \\
\hline
\rule{0pt}{2.5ex}\text{From \(Y\) to \(X\)}\rule[-1ex]{0pt}{0pt}& \text{Up} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • Using the right hand grip rule, when the switch is closed the current through the lower solenoid produces a south pole at its top.
  • So, a south pole is induced at the bottom of the solenoid connected to the spring balance (Lenz’s Law).
  • Using the right hand grip rule a second time. The current through the straight conductor goes from \(X\) to \(Y\).
  • As the two solenoids repel each other, the force on the higher solenoid counteracts its weight and the pointer moves up.

\(\Rightarrow B\)


♦♦♦ Mean mark 34%.

PHYSICS, M6 2020 HSC 33

A strong magnet of mass 0.04 kg falls 0.78 m under the action of gravity from position `X` above a hollow copper cylinder. It then travels a distance of 0.20 m through the cylinder from `Y` to `Z` before falling freely again.
 

The magnet takes 0.5 seconds to pass through the cylinder. The displacement-time graph of the magnet is shown.
 

Analyse the motion of the magnet by applying the law of conservation of energy.

Your analysis should refer to gravity and the copper cylinder, and include both qualitative and quantitative information.   (9 marks)

Show Answers Only

During the first 0.4 seconds:

  • The magnet is accelerating downwards at 9.8 m s–2 due to gravity.
  • The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.
  • Quantitatively:
  •   `Delta E_(k)= Delta U=mg Delta h=0.04 xx9.8 xx0.78=0.30576\ \text{J}`
  • Hence 0.30576 joules of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

As magnet reaches the copper cylinder:

  • Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).
  • This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 
  • Finding the magnets speed before entering the copper cylinder:
`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91\ \text{m s}^{-1}`  

 

  • Quantifying the kinetic energy loss:
`Delta E_(k)` `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2-(1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256\ \text{J}`  

 

  • Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.
  • Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.
  • Quantifying the decrease in gravitational potential energy:
  •   `Delta U=mg Delta h=0.04xx 9.8xx 0.2=0.0784`
  • Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.
Show Worked Solution

During the first 0.4 seconds:

  • The magnet is accelerating downwards at 9.8 m s–2 due to gravity.
  • The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.
  • Quantitatively:
  •   `Delta E_(k)= Delta U=mg Delta h=0.04 xx9.8 xx0.78=0.30576\ \text{J}`
  • Hence 0.30576 joules of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

As magnet reaches the copper cylinder:

  • Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).
  • This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 
  • Finding the magnets speed before entering the copper cylinder:
`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91\ \text{m s}^{-1}`  

 

  • Quantifying the kinetic energy loss:
`Delta E_(k)` `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2-(1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256\ \text{J}`  

 

  • Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.
  • Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.
  • Quantifying the decrease in gravitational potential energy:
  •   `Delta U=mg Delta h=0.04xx 9.8xx 0.2=0.0784`
  • Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.
♦ Mean mark 51%.

PHYSICS, M6 2022 HSC 32

One type of stationary exercise bike uses a pair of strong, movable magnets placed on opposite sides of a thick, aluminium flywheel to provide a torque to make it harder to pedal.
 


 

  1. Explain the principle by which these magnets make it harder to pedal.  (3 marks)
  1. The bike rider wants to increase the opposing torque on the flywheel. Justify an adjustment that could be made to the magnets to achieve this.  (3 marks)
Show Answers Only

a.    Magnetic breaking is a consequence of Lenz’s Law.

  • The changing magnetic flux through the wheel, as a result of its rotation, causes eddy currents to be induced (Faraday’s Law).
  • According to Lenz’s Law, these eddy currents produce a force which opposes the rotation of the wheel, making it more difficult to pedal.

b.    Adjustment to increase opposing torque:

  • Moving the magnets closer to the outer edge of the flywheel will increase the opposing torque on it. As the linear speed of the wheel is greater near the edge, the rate of change of flux passing through it will increase.
  • This increases the magnitude of induced emf as  `epsilon=-N(Delta theta)/(Delta t).`
  • Consequently, the opposing force and torque will increase.
  • Moving the magnets closer to the edge of the flywheel also increases the distance between the point of application of the opposing force and the axis of rotation of the wheel. As  `tau=rF`  this further increases opposing torque.
Show Worked Solution

a.    Magnetic breaking is a consequence of Lenz’s Law.

  • The changing magnetic flux through the wheel, as a result of its rotation, causes eddy currents to be induced (Faraday’s Law).
  • According to Lenz’s Law, these eddy currents produce a force which opposes the rotation of the wheel, making it more difficult to pedal.

b.    Adjustment to increase opposing torque:

  • Moving the magnets closer to the outer edge of the flywheel will increase the opposing torque on it. As the linear speed of the wheel is greater near the edge, the rate of change of flux passing through it will increase.
  • This increases the magnitude of induced emf as  `epsilon=-N(Delta theta)/(Delta t).`
  • Consequently, the opposing force and torque will increase.
  • Moving the magnets closer to the edge of the flywheel also increases the distance between the point of application of the opposing force and the axis of rotation of the wheel. As  `tau=rF`  this further increases opposing torque.

PHYSICS, M6 2019 HSC 18 MC

A circular loop of wire is connected to a battery and a lamp. The apparatus is moved from `P` to `Q` along the path shown at a constant velocity through a region containing a uniform magnetic field.
 

Which graph shows the brightness of the lamp as the apparatus moves between `P` and `Q` ?
 

Show Answers Only

`B`

Show Worked Solution
  • Initially, the current travels clockwise through the loop of wire. As it enters the magnetic field, an anticlockwise current is induced in the loop in order to induce a magnetic field out of page, opposing the external magnetic field (Lenz’s Law).
  • This decreases the net current through the loop, causing a decrease in brightness.
  • As the loop exits the magnetic field, a clockwise current is induced to create a magnetic field into the page, opposing the decrease in magnetic flux passing through it (Lenz’s Law).
  •  This increases the net current in the loop, causing an increase in brightness.

`=>B`


♦♦♦ Mean mark 25%.

PHYSICS, M6 2019 HSC 7 MC

A bar magnet is moved away from a stationary coil.

Which diagram correctly shows the direction of the induced current in the coil and the resulting magnetic polarity of the coil?
 

 

 

Show Answers Only

`D`

Show Worked Solution
  • The current through the solenoid will produce a force that opposes the magnets motion (Lenz’s law).
  • So, there will be a north pole on the right hand side of the magnet. The right hand grip rule gives the direction of current. 

`=>D`


♦ Mean mark 47%.

PHYSICS, M6 2019 HSC 5 MC

The diagram shows two coils wound around a solid iron rod. Initially the switch is closed.
 

Opening the switch will cause the galvanometer pointer to

  1. remain at a constant reading.
  2. move from a non-zero reading to a zero reading.
  3. move from a zero reading to a non-zero reading, where it remains.
  4. move from a zero reading to a non-zero reading, then back to zero.
Show Answers Only

`D`

Show Worked Solution
  • Originally, there is a constant magnetic field passing through the coil on the right due to the current through the coil on the left.
  • As there is no change in flux through the coil on the right initially the galvanometer shows a zero reading.
  • When the switch is opened, the decrease in magnetic flux through the coil on the right causes a deflection of the galvanometer to a non-zero reading.
  • The galvanometer will return to a reading of zero when the magnetic flux passing through it drops to zero. 

`=>D`

PHYSICS, M6 2020 HSC 28

A metal rod sits on a pair of parallel metal rails, 20 cm apart, that are connected by a copper wire. The rails are at 30° to the horizontal.

The apparatus is in a uniform magnetic field of 1 T which is upward, perpendicular to the table.
 

A force, `F`, is applied parallel to the rails to move the rod at a constant speed along the rails. The rod is moved a distance of 30 cm in 2.5 s.

  1. Show that the change in magnetic flux through the circuit while the rod is moving is approximately `5.2 × 10^(-2)` Wb.   (2 marks)
  1. Calculate the emf induced between the ends of the rod while it is moving, and state the direction of flow of the current in the circuit.   (2 marks)
  1. The experiment is repeated without the magnetic field.
  2. Explain why the force required to move the rod is different without the magnetic field.   (3 marks)
Show Answers Only

a.   `phi=5.2 xx10^(-2)\ \text{Wb}`

b.   `epsi=2.1 xx10^(-2)  \text{V}`

The direction of the induced current is anticlockwise as viewed from above.

c.    Repeating experiment is without the magnetic field:

  • When the magnetic field is present, the induced current results in a force acting on the rod which opposes its motion (Lenz’s Law).
  • Additionally, the force required to move the rod must also overcome the downwards gravitational force.
  • Without the magnetic field, there is no opposing force due to the induced current so the force applied only needs to overcome gravity.
  • Hence, the force required to move the rod is less without the magnetic field.
Show Worked Solution
a.
`phi` `=BA cos theta`
    `=1xx(0.3 xx0.2)xx cos 30^(@)`
    `=0.05196…`
    `~~5.2 xx10^(-2)` Wb

b.
`epsi` `=-N(Delta phi)/(t)`
    `=-1xx(5.2 xx10^(-2))/(2.5)`
    `=0.208…`
    `=2.1 xx10^(-2)\ \text{V  (V>0)`

 

  • The direction of the induced current is anticlockwise as viewed from above (Lenz’s Law).


♦ Mean mark (b) 51%.

c.    Repeating experiment is without the magnetic field:

  • When the magnetic field is present, the induced current results in a force acting on the rod which opposes its motion (Lenz’s Law).
  • Additionally, the force required to move the rod must also overcome the downwards gravitational force.
  • Without the magnetic field, there is no opposing force due to the induced current so the force applied only needs to overcome gravity.
  • Hence, the force required to move the rod is less without the magnetic field.

♦ Mean mark (c) 50%.

PHYSICS, M6 2021 HSC 31

Two identical solenoids are mounted on carts as shown. Each solenoid is connected to a galvanometer, and the solenoid on cart 1 is also connected to an open switch and a battery. The total mass of cart 1 is twice that of cart 2.
 

Explain what would be observed when the switch on cart 1 is closed. In your answer, refer to the current in each galvanometer and the initial movement of the carts.   (7 marks)

Show Answers Only
  • When the switch on cart one is closed, a direct current will flow through the cart 1 solenoid causing a deflection on `G_1.`
  • Using the right hand grip rule, this will create a magnetic field with a south pole on the right hand side of this solenoid. As this magnetic field is generated, the solenoid on cart two experiences a change in magnetic flux.
  • According to Faraday’s law, an induced emf and current will be generated in the cart 2 solenoid.
  • By Lenz’s law, this current will flow in a direction that opposes the original change in flux, causing a magnetic south pole on the left hand side of this solenoid, and a momentary deflection of `G_2` in the opposite direction to `G_1`.
  • As the two solenoids produce south poles facing each other, they will repel causing the carts to move away from each other. The law of conservation of momentum dictates that since cart 1 is double the mass of cart 2, it will have half the velocity of cart 2.
Show Worked Solution

  • When the switch on cart one is closed, a direct current will flow through the cart 1 solenoid causing a deflection on `G_1.`
  • Using the right hand grip rule, this will create a magnetic field with a south pole on the right hand side of this solenoid. As this magnetic field is generated, the solenoid on cart two experiences a change in magnetic flux.
  • According to Faraday’s law, an induced emf and current will be generated in the cart 2 solenoid.
  • By Lenz’s law, this current will flow in a direction that opposes the original change in flux, causing a magnetic south pole on the left hand side of this solenoid, and a momentary deflection of `G_2` in the opposite direction to `G_1`.
  • As the two solenoids produce south poles facing each other, they will repel causing the carts to move away from each other. The law of conservation of momentum dictates that since cart 1 is double the mass of cart 2, it will have half the velocity of cart 2.


♦ Mean mark 45%.

PHYSICS, M6 2021 HSC 10 MC

A strong magnet is moved past a copper block at a constant speed as shown.
 

What is the direction of the force acting on the copper block?

  1. To the left
  2. To the right
  3. Into the page
  4. Out of the page
Show Answers Only

`B`

Show Worked Solution
  • Eddy currents will be induced in the copper block. According to Lenz’s Law, this will produce a force that opposes the motion of the magnet.
  • This is done by minimising the relative motion between the block and the magnet, producing a force on the copper block to the right.

`=>B`


♦♦♦ Mean mark 18%.