• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Physics Question - Motors and Generators Help (1 Viewer)

Khan.Paki

Member
Joined
Jul 23, 2007
Messages
97
Gender
Male
HSC
2015
Can someone please help with the following question (all a, b, and c).

It would be greatly appreciated. Thanks :pphysics question.png
 

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,927
Location
ya mum gay
Gender
Male
HSC
2025
Uni Grad
2024
a) right hand rule
b) use t = mgd
c) since there is no motion you can equate mgd = torque to solve for magnetic field strength
 

Mr_Kap

Well-Known Member
Joined
Mar 24, 2015
Messages
1,127
Gender
Male
HSC
2015
I don't get question (a).

How can you only have the north end of the magnet, so which direction is the magnetic field?
How can I use the right hand push rule without a magnetic field direction?
 

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,927
Location
ya mum gay
Gender
Male
HSC
2025
Uni Grad
2024
I don't get question (a).

How can you only have the north end of the magnet, so which direction is the magnetic field?
How can I use the right hand push rule without a magnetic field direction?
left of the page cos magnetic field goes from North -> South
 

mohammy52

New Member
Joined
Apr 3, 2014
Messages
18
Gender
Male
HSC
N/A
The fact that there is no south pole in the diagram makes no difference -> whenever you have a single pole present in a diagram it should be automatically assumed that it's opposite pole exists somewhere (it's not possible to have a single pole exist by itself) - they just sometimes don't show it in a question, just accept it Re. how do you figure out direction of field, I understand the confusion because only a single pole is shown, but the wire on the far LHS is close enough for the field to be assumed to being perpendicular to the current. A lot of assumptions I know. But you get used to it with enough practice :)
 

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
The fact that there is no south pole in the diagram makes no difference -> whenever you have a single pole present in a diagram it should be automatically assumed that it's opposite pole exists somewhere (it's not possible to have a single pole exist by itself) - they just sometimes don't show it in a question, just accept it Re. how do you figure out direction of field, I understand the confusion because only a single pole is shown, but the wire on the far LHS is close enough for the field to be assumed to being perpendicular to the current. A lot of assumptions I know. But you get used to it with enough practice :)
It actually does matter!

This question wanted you to equate torque of ONE SIDE OF THE COIL ONLY!!

It wanted to assume that the magnetic field lines had spread out.

Notes from marking centre for that question
Weaker responses demonstrated some misunderstandings regarding torque and force. Common errors included incorrectly equating torque with force and assuming the magnetic field was constant in the equation τ = nBIAcosθ.
 
Last edited:

Mr_Kap

Well-Known Member
Joined
Mar 24, 2015
Messages
1,127
Gender
Male
HSC
2015
what year is this so i can check in the success one book for the answers?
 

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
what year is this so i can check in the success one book for the answers?
Please see attached.

It is from 2009 HSC.

Note: they used nBil * d where d is distance from axle to coil in the field.

No guarantee success one will be right btw, lol.

image.jpg
 
Last edited:

Mr_Kap

Well-Known Member
Joined
Mar 24, 2015
Messages
1,127
Gender
Male
HSC
2015
It actually does matter!

This question wanted you to equate torque of ONE SIDE OF THE COIL ONLY!!

It wanted to assume that the magnetic field lines had spread out.

Notes from marking centre for that question
Weaker responses demonstrated some misunderstandings regarding torque and force. Common errors included incorrectly equating torque with force and assuming the magnetic field was constant in the equation τ = nBIAcosθ.
"This question wanted you to equate torque of ONE SIDE OF THE COIL ONLY!!"
What does that mean? How do we know? How could the question be worded so that it wasn't only on one side.



"It wanted to assume that the magnetic field lines had spread out."

What does this mean?


" Common errors included assuming the magnetic field was constant in the equation τ = nBIAcosθ"

Isn't it? What does this mean?




And if current is moving anti-clockwise in the coil, and the magnetic field direction is left, why is the weight on side X. Shouldn;t it be on side Y?
 
Last edited:

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
"This question wanted you to equate torque of ONE SIDE OF THE COIL ONLY!!"
What does that mean? How do we know? How could the question be worded so that it wasn't only on one side.



"It wanted to assume that the magnetic field lines had spread out."

What does this mean?


" Common errors included assuming the magnetic field was constant in the equation τ = nBIAcosθ"

Isn't it? What does this mean?




And if current is moving anti-clockwise in the coil, and the magnetic field direction is left, why is the weight on side X. Shouldn;t it be on side Y?
The question says that the mass is placed on side X OR side Y, meaning that it wants you to calculate the torque created by that mass. If the same mass was placed on the other side of the coil, then the torques created by both of the masses would cancel out, and you would have no rotational moment (due to mass on side Y creating a clockwise torque, and a mass on side X would produce an anti-clockwise torque)

Regardless if there is a north or south pole located on the other side, what you have to assume that the magnetic field lines are spread out, and are moving out of the north pole magnet. This means that you can assume that the magnetic field lines are travelling to the right.

The magnetic field is not constant, because the magnetic field lines passing through side X will be greater than side Y. What you would have to use is T = fd, where f is the force created by the mass (mg), and d is the perpendicular distance from the pivot to the point of rotation.

The reason why its not on side Y is because the magnetic field lines passing through that side is almost negligible, so you assume its on side X
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top