• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

magnetic stuff (1 Viewer)

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
I dont understand these two questions...
Help would be much appreciated.
Thanks :)

I thought in Untitled-1.jpg the positive terminal should be x.
And in Untitled-2.jpg why isn't the answer B?
The first one is from 2002 STANSW.
The second one is from 2001 HSC.
 
Last edited:

Brad

Member
Joined
Oct 22, 2003
Messages
394
Location
State of mind
Q14, since in 1 the magnetic field is constantly perp to the wire when its moving.... the other one is only perpendicular at certain points.... It varies througout the movement
 

ontherun

Member
Joined
Aug 22, 2003
Messages
55
Location
Birrong
Gender
Female
HSC
2006
this is 4rom success one regarding question 14 hsc:

WITH FRICTION NEGLECTED, ROD Q EXPERIENCES ONLY A FORCE IN A VERTICAL DIRECTION, AND THUS REQUIRES NO FORCE TO MAINTAIN ITS HORIZONTAL CIRCULAR MOTION, WHITLE THE CURRENT IN ROD P PRODUCES A MAGNETIC FIELDOUT OF THE PAGE ONTO ITS LEADING SIDE, AND HENCE REQUIRES CONSTANT FORCE TO BE APPLIED TO KEEP IT ROTATING
 

shazzam

acta est fabula
Joined
Feb 4, 2004
Messages
260
Gender
Female
HSC
2004
I think Brad is right about the second q, and you can use the same principle in the first question: since the field lines are coming out of the table, they are always perpendicular to the plane of movement of the "metal rod", thus the induced emf remains constant (flux change is constant, as rod is revolving at uniform velocity). NB I'm assuming the "rod" is that thing which is in the same plane as the surface of the table...don't think it's the stick thing coming out of that cylindrical object (that may be part of the "pivot")
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
I have Success One but I don't really understand it.

For the first Q (STANSW):
I know it's constant, but I don't know why y is the positive terminal....

For the second Q (HSC):
mm I think I can see what Brad is saying... Thanks.
(the explanation in Success One is a bit different though)
 

Premus

Member
Joined
May 21, 2004
Messages
216
so for the HSC q14

If we were to draw graphs...
i would think that the graph for diagram X - would be a straight line - as you guys said the field is always perp.

What would the graph for Diagram Y look like?

Thanks
 

tempco

...
Joined
Aug 14, 2003
Messages
3,835
Gender
Male
HSC
2004
For the first question, don't the electrons move to Y? So X should be the positive end... ?
 

shazzam

acta est fabula
Joined
Feb 4, 2004
Messages
260
Gender
Female
HSC
2004
mojako said:
For the first Q (STANSW):
I know it's constant, but I don't know why y is the positive terminal....
In this particular case there are no obvious terminals we can refer to hence there's really no such thing as a POSITIVE or NEGATIVE terminal, just a direction the induced current is flowing in. If the rod changes direction midway, let's say it performs simple harmonic motion in a semi-circle, then we'd get an oscillating sine curve, ie current flowing in opposite directions at opposite times. However this is not the case, as it is in continual uniform motion.

I think it's important that they only provided ONE answer where the EMF produced is constant, had they provided TWO, one where it's "positive" and one where it's "negative", the answer would have been ambiguous.
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
it says voltage is measured from x to y and the diagram labels which one is x and y.
also in the answer in Success One, they say y is the positive one....
but if everyone thinks x is the positive then at least I'm more confident.
 

shazzam

acta est fabula
Joined
Feb 4, 2004
Messages
260
Gender
Female
HSC
2004
What do you mean Mojako, is this for diagram one or Driagram 2?
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
shazzam said:
What do you mean Mojako, is this for diagram one or Driagram 2?
the x y thing is for diagram 1
sorry for the confusion
 

FcUk

Member
Joined
Mar 30, 2004
Messages
64
isnt th first question just a constant emf coz the rod is always at the same angle to the magnetic field and it doesnt rotate or move(others than revovle) from orginal state as opposed to a loop
 

Jase

Member
Joined
Mar 7, 2004
Messages
724
Location
Behind You
Gender
Male
HSC
2004
I agree, the postive end is X, but if the answer is B, then Shazzam's right.
Physics most of the time comes down to eliminations and lines of best fits rather than a correct answer.

wait, so why the hell is the answer B??
 
Last edited:

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
oh.. guys.. really realy sorry I typed the wrong thing in my first post.
I said "And in Untitled-1.jpg the answer should be B."
I meant it's Untitled-2.jpg.
sorry once again.

so my questions are:
in 1.jpg, why isn't X the positive terminal. but this has probably been answered and the solution on the book is wrong.
in 2.jpg, why is the force on diagram Y variable

thanks.
 

Jase

Member
Joined
Mar 7, 2004
Messages
724
Location
Behind You
Gender
Male
HSC
2004
LOL! double typo!.. you had why IS the answer B, at first
And now its why ISNT the answer B.. haha that makes more sense!

Well okay on Q, what are the forces acting? Theres centripetal AND gravity. Since g is constant... the force downards is constant. However, centripetal force is also a vector directed towards the centre. So when the rod is at its lowest point, the centripetal force acts up while the gravity force acts down. then at the top, both forces at down and so on.. So the forces on Q are varying sinusoidally.
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
Jase said:
LOL! double typo!.. you had why IS the answer B, at first
And now its why ISNT the answer B.. haha that makes more sense!

Well okay on Q, what are the forces acting? Theres centripetal AND gravity. Since g is constant... the force downards is constant. However, centripetal force is also a vector directed towards the centre. So when the rod is at its lowest point, the centripetal force acts up while the gravity force acts down. then at the top, both forces at down and so on.. So the forces on Q are varying sinusoidally.
re typo:
I wrote that the answer for 2.jpg (I had 1.jpg at the time) should be B [according to my understanding], which means the same thing as why isn't the answer B. so its single typo.. but a very misleading one ;)
the correct answer is C, u can check the examiners' report to confirm

re 2.jpg and ur explanation:
dont they say it's "horizontal" circular paths in the first paragraph?
so if u combine gravity and centripetal including direction then the net force changes but the magnitude stays the same?

in diagram Y, are there any forces whatsoever caused by the presence of the magnetic field and the current?
 
Last edited:

Jase

Member
Joined
Mar 7, 2004
Messages
724
Location
Behind You
Gender
Male
HSC
2004
okay you just confused me lol.
Ah okay woops my interpretation was wrong.
that 'horizontal' is a misleading word, it should be both are moving in uniform circular motion.

Yes there is a force exerted by the field. As it moves to a horizontal position, its cutting the field at right angles, and so there's a force resisting it... meaning varying forces.
so brad's first post was essentially right..
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
hmm ok. thanks ^^

I think the circle is in the horizontal plane though.
and the force in diagram Y acts vertically. but if you also count vertical forces it varies with time, as u & Brad said.
 

CrashOveride

Active Member
Joined
Feb 18, 2004
Messages
1,488
Location
Havana
Gender
Undisclosed
HSC
2006
for the STANSW question , since the rod is always at right angles to the magnetic field, then from the sin@ in the magnetic flux equation (flux=BAsin@) there is always a max magnetic flux? and how i learnt EMF was to take the -ve derivative (so we are looking for negative gradient) of the magnetic flux curve..since magnetic flux is always at a max... derivative is zero.. ??
 
Last edited:

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

Top