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PHYSICS, M6 2024 HSC 15 MC

A uniform magnetic field is directed into the page. A conductor \(P Q\) rotates about the end \(P\) at a constant rate.
 

 
 

Which graph shows the emf induced between the ends of the conductor, \(P\) and \(Q\), as it rotates one revolution from the position shown?
 


 

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\(D\)

Show Worked Solution

→ As the conductor rotates in the magnetic field, all of the charges in the conductor will have a velocity.

→ This will produce a force on each positive and negative charge in the rod according to the equation \(F=qvB\).

→ Using the right hand rule, the direction of the force applied to each positive charge in the rod during the rotation will be towards \(P\). Similarly, the direction of force on each negative charge will be towards \(Q\). 

→ This separation of positive and negative charges to opposite ends of the conductor generates the emf.

→ As the velocity of the charges during the circular motion will remain constant, the force on the charges will be constant. Therefore, a constant emf will be produced in the conductor.

\(\Rightarrow D\)

♦♦ Mean mark 35%.

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)

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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 3-4 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 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 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?

 

 

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`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 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  \(\dfrac{1}{2}v\)  and follows an identical path.

What is the mass and charge of the second particle?

\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{Mass}\quad \rule[-1ex]{0pt}{0pt}& \quad \textit{Charge} \quad \\
\hline
\rule{0pt}{2.5ex}m\rule[-1ex]{0pt}{0pt}&q\\
\hline
\rule{0pt}{2.5ex}\frac{1}{2} m\rule[-1ex]{0pt}{0pt}& 2q\\
\hline
\rule{0pt}{2.5ex}4m\rule[-1ex]{0pt}{0pt}& q \\
\hline
\rule{0pt}{2.5ex}m\rule[-1ex]{0pt}{0pt}& \frac{1}{2}q \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

\(\rightarrow\) The centripetal force acting on the charge is given by the force it experiences due to the magnetic field:

\(F_c\) \(=F_b\)  
\(\dfrac{m v^2}{r}\) \(=q v B\)  
\(r\) \(=\dfrac{m v}{q B}\)  

 
\(\text{Given} \ B=2 B \ \text{and} \ v=\dfrac{1}{2} v:\)

\(r\) \(=\dfrac{m \frac{1}{2} v}{q(2 B)}\)  
\(r\) \(=\dfrac{m v}{4 q B}\)  

 
\(\rightarrow \text{In order for the radius to remain the same, \(\dfrac{m}{q}\) must be four}\)

\(\Rightarrow C\)


♦♦ 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?

\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}\quad \textit{Force on electron}\quad\rule[-1ex]{0pt}{0pt}& \quad \textit{Radius of path} \quad\\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}&\text{Decreases}\\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{Increases}\\
\hline
\rule{0pt}{2.5ex}\text{Decreases}\rule[-1ex]{0pt}{0pt}& \text{Decreases} \\
\hline
\rule{0pt}{2.5ex}\text{Decreases}\rule[-1ex]{0pt}{0pt}& \text{Increases} \\
\hline
\end{array}
\end{align*} 

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:

\(\dfrac{m v^2}{r}=q v B \Rightarrow r=\dfrac{m v}{q B} \Rightarrow r \propto v\)

→ Increasing the speed of the electron increases its radius.

\(\Rightarrow B\)


Mean mark 53%.

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)

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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 34

Three charged particles, `X, Y` and `Z`, travelling along straight, parallel trajectories at the same speed, enter a region in which there is a uniform magnetic field which causes them to follow the paths shown. Assume that the particles do not exert any significant force on each other.
 


 

Explain the different paths that the particles follow through the magnetic field.   (7 marks)

Show Answers Only

→ The charged particles each experience a force equal to  `F=qvB`  as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

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

 
→ As the strength of the magnetic field `(B)` and the speed `(v)` of the particles is the same, their trajectory is only affected by their mass to charge ratio `((m)/(q))`  which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both `X` and `Y` initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  `Y` has a greater radius than `X`, so `Y`’s charge to mass ratio is greater than `X`’s.

→ `Z` initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ `Z` has the same radius of motion as `X`, so the charge to mass ratios of `Z` and `X` are the same.

→ `Z` has a smaller radius than `Y`, so `Z`’s charge to mass ratio is less than that of `Y`.

Show Worked Solution

→ The charged particles each experience a force equal to  `F=qvB`  as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

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

 
→ As the strength of the magnetic field `(B)` and the speed `(v)` of the particles is the same, their trajectory is only affected by their mass to charge ratio `((m)/(q))`  which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both `X` and `Y` initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  `Y` has a greater radius than `X`, so `Y`’s charge to mass ratio is greater than `X`’s.

→ `Z` initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ `Z` has the same radius of motion as `X`, so the charge to mass ratios of `Z` and `X` are the same.

→ `Z` has a smaller radius than `Y`, so `Z`’s charge to mass ratio is less than that of `Y`.


Mean mark 60%.

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%.