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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 2020 VCE 6*

A single loop of wire moves into a uniform magnetic field \(B\) of strength 3.5 × 10\(^{-4}\) T over time \(t\) = 0.20 s from point \(\text{X}\) to point \(\text{Y}\) , as shown in the diagram below. The area \(A\) of the loop is 0.05 m\(^2\).
 

Determine the magnitude of the average induced EMF in the loop.   (2 marks)

Show Answers Only

\(8.8 \times 10^{-5}\ \text{V}\)

Show Worked Solution

\(\varepsilon=\dfrac{\Delta \Phi}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{3.5 \times 10^{-4} \times 0.05}{0.2}=8.8 \times 10^{-5}\ \text{V}\)

PHYSICS, M6 2021 VCE 6 MC

A magnet approaches a coil with six turns, as shown in the diagram below. During time interval \(\Delta t\), the magnetic flux changes by 0.05 Wb and the average induced EMF is 1.2 V.
 

Which one of the following is closest to the time interval \(\Delta t\) ?

  1. 0.04 s
  2. 0.01 s
  3. 0.25 s
  4. 0.50 s
Show Answers Only

\(C\)

Show Worked Solution
\(\varepsilon\) \(=N\dfrac{\Delta \Phi}{\Delta t}\)  
\(\Delta t\) \(=N \dfrac{\Delta \Phi}{\varepsilon}\)  
  \(=6 \times \dfrac{0.05}{1.2}\)  
  \(=0.25\ \text{s}\)  

 

\(\Rightarrow C\)

PHYSICS, M6 2022 VCE 6

Figure 6 shows a simple alternator consisting of a rectangular coil of area 0.060 m\(^{2}\) and 200 turns, rotating in a uniform magnetic field. The magnetic flux through the coil in the vertical position shown in Figure 6 is 1.2 × 10\(^{-3}\) Wb.
 

  1. Calculate the strength of the magnetic field in Figure 6 . Show your working.  (2 marks)

  1. The rectangular coil rotates at a frequency of 2.5 Hz.
  2. Calculate the average induced EMF produced in the first quarter of a turn. Begin the quarter with the coil in the vertical position shown in Figure 6.  (3 marks)

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a.    \(0.02\ \text{T}\)

b.    \(2.4\ \text{V}\)

Show Worked Solution
a.     \(\Phi\) \(=BA\)
  \(B\) \(=\dfrac{\Phi}{A}\)
    \(=\dfrac{1.2 \times 10^{-3}}{0.060}\)
    \(=0.02\ \text{T}\)

 
b.    \(\text{The time for 1 complete rotation:}\)

\(T=\dfrac{1}{f}=\dfrac{1}{2.5}=0.4\ \text{s}\)

\(\Rightarrow \text{Time for a quarter turn}\ =0.1\ \text{s}\)
 
\(\varepsilon\) \(=N\dfrac{\Delta \Phi}{t}\)  
  \(=200 \times  \dfrac{1.2 \times 10^{-3}}{0.1}\)  
  \(=2.4\ \text{V}\)  

PHYSICS, M6 2022 VCE 4

A square loop of wire connected to a resistor, \(\text{R}\), is placed close to a long wire carrying a constant current, \(I\), in the direction shown in Figure 4.

The square loop is moved three times in the following order:

  • Movement A – Starting at Position 1 in Figure 4, the square loop rotates one full rotation at a steady speed about the \(x\)-axis. The rotation causes the resistor, \(\text{R}\), to first move out of the page.
  • Movement B – The square loop is then moved at a constant speed, parallel to the current carrying wire, from Position 1 to Position 2 in Figure 4.
  • Movement C – The square loop is moved at a constant speed, perpendicular to the current carrying wire, from Position 2 to Position 3 in Figure 4.
     

Complete the table below to show the effects of each of the three movements by:

  • sketching any EMF generated in the square loop during the motion on the axes provided (scales and values are not required)
  • stating whether any induced current in the square loop is 'alternating', 'clockwise', 'anticlockwise' or has 'no current'.
Show Answers Only

Show Worked Solution

The magnetic field strength around a current carrying conductor, \(B \propto \dfrac{1}{r}\). Hence, for the movement in C, the change in magnetic field strength and therefore magnetic flux is one of inverse proportionality as depicted in the graph below.

♦ Mean mark 43%.

PHYSICS, M6 2023 VCE 5

Figure 4a 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 4a, to completely inside the magnetic field, Figure 4b.
 

  1. Calculate the average EMF produced in the loop as it moves from the position just outside the region of the field, Figure \(4 \mathrm{a}\), to the position completely within the area of the magnetic field, Figure 4b.
  2. Show your working.  (2 marks)
  1. On the small square loop in Figure 5, show the direction of the induced current as the loop moves into the area of the magnetic field.  (1 mark)
     

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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 marks)
  1. ii. Explain the behaviour of the ring.
  2.     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 2017 HSC 14 MC

The diagram shows a DC circuit containing a transformer.
 

The potential differences `V_1` and `V_2` are measured continuously for 4 s. The switch is initially closed.

At t = 2 s, the switch is opened.

Which pair of graphs shows how the potential differences `V_1` and `V_2` vary with time over the 4-second interval?
 

 

Show Answers Only

`A`

Show Worked Solution

→ When the switch is opened at t = 2 s, the current through the primary circuit and thus `V_(1)`  drops to zero.

→ This sudden decrease in magnetic flux through the secondary coil induces a temporary voltage in the secondary circuit.

`=>A`

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.    → 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.    → 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 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)

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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` J  

 

→  Hence 0.30576 J 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`  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` 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` J  

 

→  Hence 0.30576 J 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` 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` 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 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 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%.