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PHYSICS, M6 2023 HSC 16 MC

In a thought experiment, two identical parallel aluminium rods, \(X\) and \(Y\), are carrying electric currents of equal magnitude. Rod \(X\) rests on a table. Rod \(Y\) remains stationary, vertically above \(X\), as a result of the magnetic interaction. The masses of the connecting wires are negligible.
 


 

Which statement must be correct if rod \( Y\) is stationary?

  1. The magnetic force acting on \(X\) is upward.
  2. The currents through \(X\) and \(Y\) are in the same direction.
  3. The force the table exerts on \(X\) is equal and opposite to the total weight of \(X\) and \(Y\).
  4. The force the table exerts on \(X\) is equal and opposite to the force of gravity acting on \(Y\).
Show Answers Only

\(C\)

Show Worked Solution

→ As rod \(X\) and rod \(Y\) exert repulsive forces on one another, the current running through the wires must be in opposite directions and the magnetic force of \(X\) is equal and opposite to the weight force of \(Y\).

\(F_{\text{X on Y}} = m_{\text{Y}} \times g\)

→ The force the table exerts on \(X\), is equal and opposite to the magnetic force of \(Y\) on \(X\) and the weight force of \(X\).

\(F_{\text{table on X}}=F_{\text{Y on X}} + m_{\text{X}} \times g\)

→ \(F_{\text{Y on X}} = F_{\text{X on Y}}\)  as a result of Newton’s third law of motion.

→ \(F_{\text{table on X}} \) \(= F_{\text{X on Y}} + m_{\text{X}} \times g\)  
  \(= m_{\text{Y}} \times g + m_{\text{X}} \times g\)  

→ Therefore, the force the table exerts on \(X\) is equal and opposite to the weight force of \(X\) and \(Y\).

\(\Rightarrow C\)

♦♦ Mean mark 35%.

PHYSICS, M6 EQ-Bank 9 MC

Two parallel conducting rods are connected by a wire as shown and carry current `I`. They are separated by distance `d` and repel each other with a force `F`.
 

Which graph best shows how the current `I` would need to be varied with distance `d` to keep the force `F` constant?
  

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`D`

Show Worked Solution

→ Using `(F)/(l)=(mu_(0))/(2pi)(I_(1)I_(2))/(r)`  where  `I_(1)=I_(2)` and `r=d`

→ `(F)/(l)` `=(mu_(0))/(2pi)(I^2)/(d)`  
`(I^2)/(d)` `=(2piF)/(mu_(0)l)`  

 
→ In this experiment, `I` and `d` are varied while `F` is kept constant. `2pi`, `mu_(0)` and `l` are constants.

→ `(I^2)/(d)` `=k`  
`I` `=sqrt(kd)`  

 
→ Hence, the graph will be a square root function.

`=>D`

PHYSICS, M6 EQ-Bank 2 MC

A student performed an experiment using two identical, current-carrying metal rods connected to a power supply. Rod `A` was placed at different distances from Rod `B`, and the measurements on the electronic balance were recorded.
 

What is the dependent variable in this experiment?

  1. The current in `A`
  2. The length of the rods
  3. The mass recorded on the balance
  4. The distance between the two rods
Show Answers Only

`C`

Show Worked Solution

→ The dependent variable is the variable which is being measured.

→ In this experiment, the mass recorded on the balance is measured.

`=>C`

PHYSICS, M6 2015 HSC 9 MC

`P`, `Q` and `R` are straight, current-carrying conductors. They all carry currents of the same magnitude `(I)`. Conductors `P` and `Q` are fixed in place. The magnitude of the force between conductors `Q` and `R` is `F` newtons.
 

What is the net force on conductor `R` when it is in the position shown?

  1. `(F)/2` newtons to the left
  2. `(F)/2` newtons to the right
  3. `(3F)/2` newtons to the left
  4. `(3F)/2` newtons to the right
Show Answers Only

`B`

Show Worked Solution

→ As the currents in `Q` and `R` are in opposite directions, they repel. The force on `R` due to `Q` is to the right.

→ As the currents in `P` and `R` are in the same direction, they attract. The force on `R` due to `P` is to the left.

`(F)/(l)=(mu_(0))/(2pi)(I_(1)I_(2))/(r)\ \ =>\ \  F prop (1)/(r)`

→ The force on `R` due to `P` is half the magnitude of the force on `R` due to `Q`.

→ The force on `R` due to `P` is `(F)/(2)` newtons to the left.

→ `R` experiences a force of `F` newtons to the right due to `Q` and a force of `(F)/(2)` newtons to the left due to `P`.

→ The net force on `R` is `(F)/(2)` newtons to the right.

`=>B`


Mean mark 54%.

PHYSICS, M6 2017 HSC 16 MC

An AC supply is connected to a light bulb by two long parallel conductors as shown.
 

Which graph shows the variation over time of the magnetic force between the two conductors?
 

 

Show Answers Only

`D`

Show Worked Solution

→ The currents are always in opposite directions, so the two conductors always repel.

→ The magnitude of current fluctuates between zero and a maximum leading to fluctuation in magnitude of force between the conductors between zero and a maximum.

`=>D`


♦♦ Mean mark 36%.

PHYSICS, M6 2018 HSC 24b

Three parallel wires `X`, `Y` and `Z` all carry electric currents. A force of attraction is produced between `Y` and `Z`. There is zero net force on `Y`.
 

What is the magnitude and direction of the current in `X` ?   (3 marks)

Show Answers Only

50 Amps in the same direction as `Y`.

Show Worked Solution

→  Net force on `Y` = 0 (given).

→  The force of attraction between `Y` and `X` is equal to the force of attraction between `Y` and `Z`.

`(F)/(l)` `=(mu_(0))/(2pi)(I_(1)I_(2))/(r)`  
`(mu_(0))/(2pi)(I_(Y)I_(Z))/(0.3)` `=(mu_(0))/(2pi)(I_(X)I_(Y))/(0.75)`  
`(I_(Y)I_(Z))/(0.3)` `=(I_(X)I_(Y))/(0.75)`  
`(20 xx20)/(0.3)` `=(20I_(X))/(0.75)`  
`I_(X)` `=50\ text{A}`  

 
→ The direction of current in `X` is the same as the direction of current in `Y`, as there is a force of attraction between `X` and `Y`.

PHYSICS M6 2022 HSC 15 MC

Two wires separated by a distance, `d`, carry equal electric currents producing a magnetic force between them.
  


  

The separation between the wires is increased to `4 d` and the current in each wire is doubled.

What happens to the magnetic force between the wires, compared to the original force?

  1. It does not change.
  2. It increases by a factor of 4 .
  3. It decreases by a factor of 4 .
  4. It decreases by a factor of 8 .
Show Answers Only

`A`

Show Worked Solution

The force per unit length between the wires:

`(F)/(l)=(mu_(0))/(2pi)(I_(1)I_(2))/(r)`
 

By increasing the distance by a factor of four, and doubling both currents, the new force per unit length is given by:

`(F)/(l)` `=(mu_(0))/(2pi)(2I_(1)2I_(2))/(4r)`  
  `=(mu_(0))/(2pi)(I_(1)I_(2))/(r)`  

 
`=>A`

PHYSICS, M6 2020 HSC 14 MC

Two parallel wires, `X` and `Y`, each carry a current `I`.
 

The magnitude and direction of the force on wire `Y` are represented by the vector `F`.

The current in wire `Y` is then doubled and its direction is reversed. The current in wire `X` remains unchanged.

Which vector arrow represents the force on wire `X` after the change to the current in wire `Y`?
 

Show Answers Only

`C`

Show Worked Solution

Reversal of current in `Y`  → attractive force between wires.

Doubling of the magnitude of current in `Y` → doubling of the magnitude of force (Ampere’s Law)

`=>C`


♦ Mean mark 43%.

PHYSICS, M6 2021 HSC 17 MC

Two long, parallel conductors `X` and `Y` are connected to a light bulb and an AC power supply. The conductors are suspended horizontally from fixed points using sensitive spring balances. `X` is positioned directly below `Y`.
 


 

Which statement correctly compares the forces measured by the spring balances?

  1. The forces measured on `X` and `Y` will always be equal.
  2. The force measured on `Y` will be greater than or equal to that on `X`.
  3. The force measured on `X` will be greater than or equal to that on `Y`.
  4. There will be a continuous reversal of which measured force is greater.
Show Answers Only

`C`

Show Worked Solution

Current in wires in opposite direction → The two conductors will always repel (Ampere’s Law)

Resulting forces:

→ upwards force on Y which partially cancels its weight, decreasing the net downwards force acting on it

→ downwards force on X which increases the net downwards force acting on it.

`=>C`


♦ Mean mark 23%.