• 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

2018 HSC Q20 (1 Viewer)

askit

Member
Joined
Oct 10, 2023
Messages
66
Gender
Male
HSC
2024
I'm not sure how to go about this question:
 

liamkk112

Well-Known Member
Joined
Mar 26, 2022
Messages
1,059
Gender
Female
HSC
2023
firstly, a galvanometer measures current, so the pointer will deflect in the direction of current.

now, when X rotates, there is relative motion between the magnet inside the galvanometer and the conductive wire, so by faraday's law there will be some induced current in the circuit. according to the diagram and by lenz's law, the anticlockwise rotation of galvanometer X means that we would like to push the galvanometer back in the other direction, using induced currents. using right hand rule, we can thus see that the current induced would need to be from the positive terminal of X to the negative of Y. hence, the current will flow from negative to positive in Y; this would mean that the pointer is deflected towards the left. so, C and D are the only remaining options.

now a motor is just something that converts electrical energy to mechanical energy. in essence this is what's happening; the current through Y is being converted to kinetic energy to oppose the motion of X. this is kind of confusing on purpose, because it seems like Y is acting like a generator, converting the kinetic energy of X into some current; however, it is X that is the generator, as the kinetic energy of the rotation is turned into the current through induction, and then Y converts this current back into kinetic energy through opposing the motion of X.
 

askit

Member
Joined
Oct 10, 2023
Messages
66
Gender
Male
HSC
2024
firstly, a galvanometer measures current, so the pointer will deflect in the direction of current.

now, when X rotates, there is relative motion between the magnet inside the galvanometer and the conductive wire, so by faraday's law there will be some induced current in the circuit. according to the diagram and by lenz's law, the anticlockwise rotation of galvanometer X means that we would like to push the galvanometer back in the other direction, using induced currents. using right hand rule, we can thus see that the current induced would need to be from the positive terminal of X to the negative of Y. hence, the current will flow from negative to positive in Y; this would mean that the pointer is deflected towards the left. so, C and D are the only remaining options.

now a motor is just something that converts electrical energy to mechanical energy. in essence this is what's happening; the current through Y is being converted to kinetic energy to oppose the motion of X. this is kind of confusing on purpose, because it seems like Y is acting like a generator, converting the kinetic energy of X into some current; however, it is X that is the generator, as the kinetic energy of the rotation is turned into the current through induction, and then Y converts this current back into kinetic energy through opposing the motion of X.
heya sorry if I'm being really stupid rn but I'm not seeing this: using induced currents. using right hand rule, we can thus see that the current induced would need to be from the positive terminal of X to the negative of Y. hence, the current will flow from negative to positive in Y; this would mean that the pointer is deflected towards the left. so, C and D are the only remaining options.

Are we talking about the coiled wire inside the galvanometer?
 

liamkk112

Well-Known Member
Joined
Mar 26, 2022
Messages
1,059
Gender
Female
HSC
2023
heya sorry if I'm being really stupid rn but I'm not seeing this: using induced currents. using right hand rule, we can thus see that the current induced would need to be from the positive terminal of X to the negative of Y. hence, the current will flow from negative to positive in Y; this would mean that the pointer is deflected towards the left. so, C and D are the only remaining options.

Are we talking about the coiled wire inside the galvanometer?
yeah, if you like the wire connecting the two meters continues inside of the meters. fyi, galvanometers are out of syllabus, so you probably won't be asked something like this anymore (or at least, they probably won't assume that you know how a galvanometer works / what it is)
 

wizzkids

Well-Known Member
Joined
Jul 13, 2016
Messages
339
Gender
Undisclosed
HSC
1998
The correct answer is (C).
Galvanometer 'X' is normally stationary. In normal operation if a positive potential is applied to the + terminal of a galvanometer the armature/needle deflects clockwise to the right (an application of the motor effect).
Now if instead the body of galvanometer 'X' is rotated anticlockwise, the armature, which is free to rotate but is suspended by clock-spring wires, is now behaving as a generator. When the magnet assembly (stator) rotates anticlockwise the armature rotates clockwise relative to the stator and a positive potential will appear at the + terminal of galvanometer 'X'. Hence the negative terminal of galvanometer 'Y' receives this positive potential and a current flows from right to left. Its armature/needle deflects anticlockwise to the left (an application of the motor effect).
So the needle of 'Y' rotates to the left, because a current has flowed from from right to left, which agrees with option (C).
I hope you can follow the logic.
 
Last edited:

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

Top