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PHYSICS, M6 2024 HSC 10 MC

A rod carrying a current, \(I\), placed in a uniform magnetic field as shown, experiences a force \(F\).
 

How many degrees must the rod be rotated clockwise so that it experiences a force \(\dfrac{F}{2}\)?

  1. 30°
  2. 45°
  3. 60°
  4. 90°
Show Answers Only

\(C\)

Show Worked Solution

→ The force experienced by a current carrying conductor in a magnetic field is given by \(F=lIB\sin \theta\), where \(\sin \theta\) refers to the angle between the direction of the current through the conductor and the magnetic field lines.

→ For the rod to experience a force of \(\dfrac{F}{2}\), the value of  \(\sin \theta = \dfrac{1}{2}\).

\(\sin \theta=\dfrac{1}{2}\ \ \Rightarrow \ \theta=\sin^{-1}\bigg(\dfrac{1}{2}\bigg)=30^{\circ}\) 

→ As the angle between the current carrying conductor and magnetic field lines originally is \(90^{\circ}\), the rod must be rotated by \(60^{\circ}\).

\(\Rightarrow C\) 

♦♦ Mean mark 31%.

PHYSICS, M6 2021 VCE 2

A schematic side view of one design of an audio loudspeaker is shown in Figure 2. It uses a current carrying coil that interacts with permanent magnets to create sound by moving a cone in and out.
 

Figure 3 shows a schematic view of the loudspeaker from the position of the eye shown in Figure 2. The direction of the current is clockwise, as shown.
 

   

  1. Draw four magnetic field lines on Figure 3, showing the direction of each field line using an arrow.  (1 mark)

  2. Which one of the following gives the direction of the force acting on the current carrying coil shown in Figure 3?  (1 mark)
A. left B. right
C. up the page D. down the page
E. into the page F. out of the page
  1. The current carrying coil has a radius of 5.0 cm and 20 turns of wire, and it carries a clockwise current \((I)\) of 2.0 A. Its magnetic field strength \((B)\) is 200 mT.
  2. Calculate the magnitude of the force, \(F\), acting on the current carrying coil. Show your working.  (2 marks)

Show Answers Only

a.   
         

b.    \(E\)

c.    \(2.51\ \text{N}\)

Show Worked Solution

a.    Magnetic fields run from the north pole to the south pole.
 

   

 

b.    Apply the right-hand rule:

→ Thumb to the right and fingers down, the force on the current carrying coil must be into the page.

\(\Rightarrow E\)
 

c.    \(l= 2\pi \times 5 \times 10^{-2} = 0.1\pi\)

\(\therefore F=nlIB=20 \times 0.1\pi \times 2 \times 0.2=2.51\ \text{N}\)

♦♦♦ Mean mark 24%.

PHYSICS, M6 2022 VCE 1

Figure 1 shows four positions (1, 2, 3 and 4) of the coil of a single-turn, simple DC motor. The coil is turning in a uniform magnetic field that is parallel to the plane of the coil when the coil is in Position 1, as shown.

When the motor is operating, the coil rotates about the axis through the middle of sides \(L M\) and \(N K\) in the direction indicated. The coil is attached to a commutator. Current for the motor is passed to the commutator by brushes that are not shown in Figure 1.
 

  1. When the coil is in Position 1, in which direction is the current flowing in the side \(K L-\) from \(K\) to \(L\) or from \(L\) to \(K\)? Justify your answer.  (2 marks)

  1. When the coil is in Position 3, in which direction is the current flowing in the side \(KL-\) from \(K\) to \(L\) or
    from \(L\) to \(K\)?      (1 marks)

  1. The side \(K L\) of the coil has a length of 0.10 m and experiences a magnetic force of 0.15 N due to the magnetic field, which has a magnitude of 0.5 T.

    Calculate the magnitude of the current in the coil.  (2 marks)

Show Answers Only

a.    \(K\) to \(L\)

b.    \(L\) to \(K\)

c.    \(3\ \text{A}\)

Show Worked Solution

a.    Using the right hand rule:

→ Palm faces up (force), fingers to the right (magnetic field), and the thumb faces out of the page.

→ Therefore the current must run from \(K\) to \(L\).
 

b.    Using the right hand rule:

→ Palm now faces down the page (force), fingers still to the right.

→ The current will run from \(L\) to \(K\) (thumb).

 

c.    \(F\) \(=lIB\)
  \(I\) \(=\dfrac{F}{lB}\)
    \(=\dfrac{0.15}{0.1 \times 0.5}\)
    \(=3\ \text{A}\)

PHYSICS, M6 2022 VCE 1 MC

A single loop of wire carries a current, \(I\), as shown in the diagram below.
 

Which one of the following best describes the direction of the magnetic field at the centre of the circle, \(\text{C}\), which is produced by the current carrying wire?

  1. to the left
  2. to the right
  3. into the page
  4. out of the page
Show Answers Only

\(C\)

Show Worked Solution

→ Right hand grip rule: fingers curl in a clockwise direction and the thumb points into the page.

\(\Rightarrow C\)

PHYSICS, M6 2023 HSC 10 MC

Figure \(\text{I}\) shows a current flowing through a loop of wire that is in a uniform magnetic field.
 

 

 

The loop is then rotated to the position shown in Figure \(\text{II}\).

The magnitude of the force on the side \(X Y\) and the magnitude of the torque on the loop in Figure \(\text{II}\) are compared to those in Figure \(\text{I}\).

Which row of the table correctly describes the comparison?

\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}\quad \textit{Force}\quad \rule[-1ex]{0pt}{0pt}&\quad \textit{Torque}\quad  \\
\hline
\rule{0pt}{2.5ex}\text{I > II}\rule[-1ex]{0pt}{0pt}&\text{I = II}\\
\hline
\rule{0pt}{2.5ex}\text{I > II}\rule[-1ex]{0pt}{0pt}& \text{I > II}\\
\hline
\rule{0pt}{2.5ex}\text{I = II}\rule[-1ex]{0pt}{0pt}& \text{I = II} \\
\hline
\rule{0pt}{2.5ex}\text{I = II}\rule[-1ex]{0pt}{0pt}& \text{I > II} \\
\hline
\end{array}
\end{align*} 

Show Answers Only

\(D\)

Show Worked Solution

→ The magnitude of the force on side \(XY\) can be calculated by  \( F= lIB \sin \theta\)

→ As side \( XY\) is always perpendicular to the magnetic field,  \( \theta = 90^{\circ} \) and Force \(\text{I}\) = Force \(\text{II}\)

→ The magnitude of the torque on side \(XY\) can be calculated by  \( \tau=rF\sin\theta\)

→ In Figure \(\text{I}\), \( \theta= 90^{\circ} \) and in Figure \(\text{II}\), \( \theta = 0^{\circ}  \) → Torque \(\text{I}\) > Torque \(\text{II}\)

\(\Rightarrow  D\)

♦ Mean mark 52%.

PHYSICS, M6 EQ-Bank 21

A current of 4.0 A flows in a wire that is placed in a magnetic field of 0.75 T. The wire is 0.80 m long and is at an angle of 45° to the field.
 


 

Calculate the force on the wire.  (2 marks)

Show Answers Only

`1.7\ text{N out of the page.}`

Show Worked Solution
`F` `=lIB sin theta`  
  `=0.8 xx4.0 xx0.75 xx sin 45^(@)`  
  `=1.7\ text{N out of the page.}`  

PHYSICS, M6 2015 HSC 22

The diagram represents a simple DC motor. A current of 1.0 A flows through a square loop `ABCD` with 5 cm sides in a magnetic field of 0.01 T.
 

  1. Determine the force acting on section `AB` and the force acting on section `BC` due to the magnetic field, when the loop is in the position shown.   (3 marks)
  1. How is the direction of the torque maintained as the loop rotates 360° from the position shown?   (2 marks)
Show Answers Only

a.    `F_(AB)=5xx10^(-4)  text{N into the page.}`

`F_(BC)=0  text{N}`

b.  The split ring conductor reverses the direction of current.

→ This switches the direction of force on sides `AB` and `CD` of the coil every half turn, ensuring torque remains in the same direction.

Show Worked Solution
a.    `F_(AB)` `=BIℓsin theta`
    `=0.01 xx1.0 xx0.05 xx sin 90^(@)`
    `=5xx10^(-4)  text{N into the page.}`
     
  `F_(BC)` `=0.01 xx1.0 xx0.05 xx sin 0^(@)`
    `=0\ text{N}`

 

b.  The split ring conductor reverses the direction of current.

→ This switches the direction of force on sides `AB` and `CD` of the coil every half turn, ensuring torque remains in the same direction.


Mean mark (b) 57%.

PHYSICS, M6 2015 HSC 7 MC

A current-carrying wire is placed perpendicular to a magnetic field.

Which graph correctly shows the relationship between magnetic field strength `(B)` and current `(I)` if the force is to remain constant?
 

 

Show Answers Only

`C`

Show Worked Solution

`F=IlB sin theta \ \ =>\ \ I=(F)/(lB sin theta)`

→  If `F` is constant, `I prop (1)/(B)`

→ i.e. there is an inverse relationship between `I` and `B`.

`=>C`


♦ Mean mark 43%.

PHYSICS, M6 2016 HSC 16 MC

The cone of a speaker is pushed so that the coil moves in the direction shown.
 

Which row of the table correctly identifies the behaviour of the speaker and the direction of the current through the conductor?

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex}\textit{} & \textit{} \\
\textit{}\rule[1ex]{0pt}{0pt}& \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
\rule{0pt}{1.5ex}\textit{The speaker} & \textit{The direction of the} \\
\textit{behaves like a ...}\rule[1ex]{0pt}{0pt}& \textit{current is from ...} \\
\hline
\rule{0pt}{2.5ex}\text{generator}\rule[-1ex]{0pt}{0pt}&\text{\(X\) to \(Y\)}\\
\hline
\rule{0pt}{2.5ex}\text{generator}\rule[-1ex]{0pt}{0pt}& \text{\(Y\) to \(X\)}\\
\hline
\rule{0pt}{2.5ex}\text{motor}\rule[-1ex]{0pt}{0pt}& \text{\(X\) to \(Y\)} \\
\hline
\rule{0pt}{2.5ex}\text{motor}\rule[-1ex]{0pt}{0pt}& \text{\(Y\) to \(X\)} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

→ Motion of the coil causes the induction of a current causing the speaker to behave like a generator.

→ A magnetic north pole is induced on the right hand side of the speaker coil in order to oppose the motion of the coil away from the magnet by creating an attractive force between the coil and the magnet (Lenz’s Law).

→ Using the right hand grip rule the direction of induced current is from \(X\)  to \(Y\).

\(\Rightarrow A\)


♦♦♦ Mean mark 25%.

PHYSICS, M6 2017 HSC 22b

A coil consisting of 15 turns is placed in a uniform 0.2 T magnetic field between two magnets. A current of 7.0 amperes flows in the direction shown.
 

Calculate the magnitude and direction of the torque produced by the side `BC` of the 15-turn coil.   (3 marks)

Show Answers Only

0.05 Nm into the page

Show Worked Solution
  `F_(BC)` `=BIl sin theta`
    `=0.2 xx7xx0.08 xx sin 90^(@)`
    `=0.112  text{N}`

 
`text{Total force (15 turns)}  =0.112 xx15=1.68\ text{N}`

`:.tau` `=rF sin theta`
  `=1.68 xx0.03 xx sin90^(@)`
  `=0.05  text{Nm into the page}`

Mean mark 58%.

PHYSICS, M6 2017 HSC 17 MC

A magnet rests on an electronic balance. A rigid copper rod runs horizontally through the magnet, at right angles to the magnetic field. The rod is anchored so that it cannot move.
 

Which expression can be used to calculate the balance reading when the switch is closed?

  1. `0.200\ text{kg} +BIl`
  2. `0.200\ text{kg} +(BIl)/9.8`
  3. `0.200\ text{kg} -BIl`
  4. `0.200\ text{kg} -(BIl)/9.8`
Show Answers Only

`B`

Show Worked Solution

→ Using the right hand palm rule, the force on the conductor due to the magnet is up.

→ By Newton’s Third Law, an equal force is exerted downwards on the magnet, adding to the reading on the scale.

→ Since the scale shows mass, not force, `F=BIl`  is divided by 9.8 to convert to a mass reading.

`=>B`


♦♦♦ Mean mark 27%.

PHYSICS, M6 2017 HSC 13 MC

A triangular piece of wire is placed in a magnetic field as shown.
 

When current \(I\) is supplied as shown, how does the wire move?

\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{Axis of rotation}\rule[-1ex]{0pt}{0pt}& \textit{Direction of movement} \\
\hline
\rule{0pt}{2.5ex}YZ\rule[-1ex]{0pt}{0pt}&Q\ \text{into page}\\
\hline
\rule{0pt}{2.5ex}YZ\rule[-1ex]{0pt}{0pt}& Q\ \text{out of page}\\
\hline
\rule{0pt}{2.5ex}WX\rule[-1ex]{0pt}{0pt}&R\  \text{into page} \\
\hline
\rule{0pt}{2.5ex}WX\rule[-1ex]{0pt}{0pt}& R\ \text{out of page} \\
\hline
\end{array}
\end{align*} 

Show Answers Only

\(C\)

Show Worked Solution

→ Using the right hand palm rule, there will be a force into the page acting on \(QR\) and a force out the page acting on \(QP\). 

→ The wire will rotate around \(WX\) with \(R\)moving into the page.

\(\Rightarrow C\)

PHYSICS, M6 2019 HSC 28

A metal loop, `WXYZ` is connected to a battery and placed in a uniform magnetic field. A current flows through the loop in the direction shown.
 

The loop is then allowed to rotate by 90° about the axis `PQ`.

Compare the forces acting on `WX` and `XY` before and after this rotation.   (3 marks)

Show Answers Only

Originally:

→ The current through  `XY`  is parallel to the magnetic field lines so it experiences no force.

→ `WX` experiences a force given by `F=BIl`  into the page.

After a `90^(@)` rotation:

→ `XY` experiences a force to the right.

→ The magnitude and direction of the force on `WX` remains the same.

Show Worked Solution

Originally:

→ The current through  `XY`  is parallel to the magnetic field lines so it experiences no force.

→ `WX` experiences a force given by `F=BIl`  into the page.

After a `90^(@)` rotation:

→ `XY` experiences a force to the right.

→ The magnitude and direction of the force on `WX` remains the same.


♦ Mean mark 46%.

PHYSICS, M6 2019 HSC 17 MC

A straight current-carrying conductor, `QR`, is connected to a battery and a variable resistor. `QR` is enclosed in an evacuated chamber with a fluorescent screen at one end.
 


 

A cathode ray enters the chamber directly above `Q`, initially travelling parallel to `QR`. It passes through the chamber and strikes the fluorescent screen causing a bright spot.

Which direction will this spot move towards if the resistance is increased?

  1. `W`
  2. `X`
  3. `Y`
  4. `Z`
Show Answers Only

`D`

Show Worked Solution

Using the right hand grip rule, the current through creates a magnetic field to the right above it. Using the right hand palm rule, this creates an upwards force on the cathode ray, which is made up of electrons.

Increasing the resistance

→ decreased current through `QR`

→ decreased upwards force on the cathode ray

→ bright spot moves downwards towards `Z`.

`=>D`


♦ Mean mark 36%.

PHYSICS, M6 2021 HSC 27

A student is considering how to levitate a thin metal rod in a strong magnetic field of 1.2 T. The current flowing through the rod will be 2.3 A.

  1. Use a labelled diagram to show a suitable orientation of the current and the magnetic field to achieve this outcome. Include relevant forces in your diagram.   (3 marks)
     
     
  1. Explain why the maximum mass per unit length of the rod cannot exceed 0.282 kg m\(^{-1}\). Support your answer with a calculation.   (3 marks)
Show Answers Only

a.


 

b.   The upwards magnetic force → \(F=IlB\, \sin \theta\)

Levitation will occur when

\(W\) \(=F_B\)  
\(mg\) \(=IlB\, \sin \theta\)  
\(\dfrac{m}{l}\) \(=\dfrac{IB\, \sin \theta}{g}\)  
  \(=\dfrac{2.3 \times 1.2 \times \sin\, 90^{\circ}}{9.8}\)  
  \(=0.282\ \text{kg m}^{-1}\)  

 
If the mass per unit length exceeds this, the rod would not levitate.

Show Worked Solution

a.

Note: a different alignment with the magnetic force coming out of the page and the current moving left is also correct.
 

b.   The upwards magnetic force → \(F=IlB\, \sin \theta\)

Levitation will occur when

\(W\) \(=F_B\)  
\(mg\) \(=IlB\, \sin \theta\)  
\(\dfrac{m}{l}\) \(=\dfrac{IB\, \sin \theta}{g}\)  
  \(=\dfrac{2.3 \times 1.2 \times \sin\, 90^{\circ}}{9.8}\)  
  \(=0.282\ \text{kg m}^{-1}\)  

 
If the mass per unit length exceeds this, the rod would not levitate.