HSC Physics Question Thread

[Happy Physics Land]

Thank you so much to you both for the comprehensive responses! Although HPL, where did you get the equation you used for Q2 from? I don't seem to be able to find it on the data sheet D: And also, I was revising some past papers and I forgot the relationship between the voltage in DC motors and the resulting speed current. The question I"m having trouble with is:

A 12V DC motor has the input voltage increased from 0V to 12V. What happens to the speed and current?
a) The rotor speed is constant but the current in the motor increases
b) The rotor speed increases and the current in the motor increases
c) The rotor speed is constant but the current in the motor decreases
d) The rotor speed increases and the current in the motor decreases

The answer is b and I was wondering why? Thank you!

(Sorry for the amount of questions I have D:)

Neutron

Ok so jamon adopted the kepler's third law approach whereas I adopted the gravitational acceleration approach. Of course, both are viable, since we all got the same answer, and in both cases, the mass of the orbiting object is a red herring.

So to derive the equation I used:
We all know that gravitational force constitutes to weight
weight = mg
Newton's law of gravitational force: F(g) = GMm/r^2 (where F(g) means gravitational force of attraction)
so if we equate the two values, we get:
GMm/r^2 = mg
Cancelling m from both sides, we get g = GM/r^2 which is where I got my equation from

[jamonwindeyer]

Ok so jamon adopted the kepler's third law approach whereas I adopted the gravitational acceleration approach. Of course, both are viable, since we all got the same answer, and in both cases, the mass of the orbiting object is a red herring.

So to derive the equation I used:
We all know that gravitational force constitutes to weight
weight = mg
Newton's law of gravitational force: F(g) = GMm/r^2 (where F(g) means gravitational force of attraction)
so if we equate the two values, we get:
GMm/r^2 = mg
Cancelling m from both sides, we get g = GM/r^2 which is where I got my equation from

I think we actually did the same thing! I just equated straight to Newton's 2nd Law whereas you included a contextual formula for weight force, which is probably a little more correct. Tertiary Physics has me taking shortcuts 

[smiley2101]

Hi! If a projectile has a time of flight of 7.5 seconds and a range of 1200m then how would you work out the maximum height? Thank you!!

[Happy Physics Land]

Hi! If a projectile has a time of flight of 7.5 seconds and a range of 1200m then how would you work out the maximum height? Thank you!!

Hey Smiley!

This question sounds a little vague to me. If this is the full question, my solution will be shown below. If you have the full question, I would gladly want to have a look at it!

Ok so in projectile motion, the object is only acted upon by gravity, meaning that horizontal velocity stays constant (i.e. initial horizontal velocity = final horizontal velocity)
Let u(x) = initial horizontal velocity
Range = u(x) x time
1200 = u(x) x 7.5
u(x) = 160m/s
So Im assuming here that you are launching your projectile from ground-level and it is going to trace out a parabolic trajectory. In this case, I would really need the launching angle to be able to figure out the initial vertical velocity. So now we have figured out the horizontal velocity to be 160m/s, you can just use your trigonometry to figure out the initial vertical velocity. Because we know that at max height, vertical velocity = 0 (i.e. v = 0), apply the v = u + at formula to figure the time at which max height occurs. Then apply the formula y = ut + 1/2 at^2 to figure out the max height.

Hope my answer helps, if you have any further concerns please dont hesitate to ask!

Best Regards
Happy Physics Land

[Meckenza]

Hello!

I am confused about the answer to the following question:

A copper ring is suspended near to a fixed solenoid S as shown.
When the switch is closed a large current flows through S. As a result, the ring R
Answer: is repelled towards S as long as the switch is closed.

Any help would be much appreciated. Thanks!

[FallonXay]

Heyy. I'm just looking over a practice exam and the answers say that there is maximum torque on a current currying loops occurs when the angle between the loop's plane and magnetic field vector is 90 degrees.
Why does maximum torque on a current carrying loop occur when the angle between the loop's plane and magnetic field vector is 90 degrees if using the formula T=nBIAcosTheta, cos90=0 and cos 0=1?

[Happy Physics Land]

Hello!

I am confused about the answer to the following question:

A copper ring is suspended near to a fixed solenoid S as shown.
When the switch is closed a large current flows through S. As a result, the ring R
Answer: is repelled towards S as long as the switch is closed.

Any help would be much appreciated. Thanks!

Hey Meckenza:

Thanks for posting! Definitely an interesting question! (quite tricky actually)

Let's have a look at this together. Whenever I approach this type of question, I just pick out the most useful information and briefly state them. So in this case:
- Copper ring suspended
- Current flows through solenoid when circuit is closed
- In the circuit, current flows clockwise (electric current flows from positive to negative terminal, and you can see the direction of current in the circuit by looking at the battery/power source)
- Copper is a conductor, and we can deduce that the core of the solenoid is also a conductor (large current means that the core must have enabled a larger current to be produced through concentrating the magnetic field of the solenoid)

Ok so I have done a deep analysis of the question. Now its time to tackle the question! Look at the direction in which the current in the circuit travels, and looking at how the solenoid is wound onto the core, we can tell that the current direction is out of the page when below the core and into the page when above the core. Using other right hand coil rule, we can effective determine that the North pole is on the left hand side of the solenoid and South is on the right hand side of the solenoid. So because of this, since the magnetic field lines would be travelling from North to South, you would get more field lines out of the page when you observe through the ring. (I have attached a diagram below to help explaining what Im trying to say).



Because there is a change in the copper ring's magnetic flux, then according to Lenz's law there will be a current flowing in the ring in such a direction that it opposes this change in flux. So in order to minimise the amount of out of page magnetic field lines, the current in the ring would want to flow anticlockwise and again by applying the right hand coil rule, we can ascertain that North is on the left hand side of the ring and South is on the right hand side. Therefore the ring would be attracted to the solenoid. Really hard question actually!



I hope my answer helps you in some sort of way haha. If you have any questions please dont hesitate to ask!

Best Regards
Happy Physics Land

[Happy Physics Land]

Heyy. I'm just looking over a practice exam and the answers say that there is maximum torque on a current currying loops occurs when the angle between the loop's plane and magnetic field vector is 90 degrees.
Why does maximum torque on a current carrying loop occur when the angle between the loop's plane and magnetic field vector is 90 degrees if using the formula T=nBIAcosTheta, cos90=0 and cos 0=1?

Hey FallonXay!

The loop's plane should be parallel to the magnetic field in order to produce the maximum torque, just like what you have correctly stated, cos 0 = 1. But the answer from your practise exam could perhaps be saying that the current through loop is at 90 degrees to the magnetic field? I highly doubt the correctness of the exam answer here. If this is a question with an image, would you mind to send the image as well? That may cause the answer to be different. But yeah the loop has to be parallel to magnetic field for maximum torque.

Best Regards
Happy Physics Land

[Meckenza]

sorry, would you be able to clarify why the current flows anticlockwise in the ring? I don't quite understand that part. Also, If the ring were attracted to the solenoid wouldn't the answer be A - I have attached an image of the full question -? (The answer page says the correct answer is B).

Thanks!

[Happy Physics Land]

sorry, would you be able to clarify why the current flows anticlockwise in the ring? I don't quite understand that part. Also, If the ring were attracted to the solenoid wouldn't the answer be A - I have attached an image of the full question -? (The answer page says the correct answer is B).

Thanks!

Hello Meckenza:

What makes this question confusing is that it has the option "attracted towards" and "repelled towards". I dont quite understand how the ring can be repelled "towards" the solenoid, because usually what happens is the ring is either "attracted towards" or "repelled away". I do remain dubious of the answer, I will try to find out the answer for you to the best of my ability.

In regards to the anticlockwise flow of current inside the ring, it is because when you look through the ring (hypothetically) you can see magnetic field lines coming out of page. So originally there wasnt any magnetic field lines travelling through the ring, but now because we established a complete circuit by closing the switch, we have made magnetic field lines that will come out of the page. So Lenz's law states that the current in the ring will flow in such a direction that it will oppose the change in magnetic flux. Since in this case the change in flux is a result of more out of page magnetic field lines, then consequently the current in the ring will flow in such a direction that it will produce into the page magnetic field lines that oppose this increase in out of page magnetic field lines. So if we apply the right hand grip rule, with our thumb pointing away from us (i.e.into the page), you can see that your fingers will curl in an anti-clockwise direction.

In regards to this question I will go find out the answer for you and I will get back to you soon!

Best Regards
Happy Physics Land

[Meckenza]

ok thank you so much!

[Maz]

hi,
we have a test on 'the standard model' coming up, and i was wandering if you could please help me with this question?
when a muon and an anti-muon collide they can annihilate each other and release their mass-energy as 2 photons. assuming that these two photons are identical,
a) what will each of their energies be
b) what wavelength will they have
c) why does there need to be 2 photons produced and not just one?
d) in what directions would they have to travel relative to each other and why?
e) in what part of the electromagnetic spectrum are they located?
i don't have any answers to these so i even the ones that i have attempted, i don't know if the answer is right or wrong

thankyou soooo much in advance

[Happy Physics Land]

ok thank you so much!

Hey Meckenza!

Sorry for letting you wait for so long. It just seemed like I got my directions wrong. So when I said using right hand grip rule we get North being on the left hand side of the solenoid and South being on the right hand side, I decided that magnetic field line is going to go towards the right. But I made a conceptual error here. This north and south is OUTSIDE the solenoid. Inside the solenoid its different. Inside the solenoid the South pole is on the left and North pole is on the right, meaning that flux is actually going left. So according to Lenz's law, the current inside the ring should be flowing in a direction that it opposes the flux change to the left (i.e. the current will flow in side the ring in such a direction that it produces a flux going to the right). So if you use your right hand grip rule the current is actually flowing anticlockwise inside the ring. Hence to the left of the ring would be South and to the right of the ring would be North. Hence it repels away from the solenoid (the wording in the answer is really bad, there is absolutely no such thing as "repel towards", only "repel away from").

Sorry for all my previous tedious explanations, I hope you understand this explanation of mine haha.

Best Regards
Happy Physics Land

[Meckenza]

Hey Meckenza!

Sorry for letting you wait for so long. It just seemed like I got my directions wrong. So when I said using right hand grip rule we get North being on the left hand side of the solenoid and South being on the right hand side, I decided that magnetic field line is going to go towards the right. But I made a conceptual error here. This north and south is OUTSIDE the solenoid. Inside the solenoid its different. Inside the solenoid the South pole is on the left and North pole is on the right, meaning that flux is actually going left. So according to Lenz's law, the current inside the ring should be flowing in a direction that it opposes the flux change to the left (i.e. the current will flow in side the ring in such a direction that it produces a flux going to the right). So if you use your right hand grip rule the current is actually flowing anticlockwise inside the ring. Hence to the left of the ring would be South and to the right of the ring would be North. Hence it repels away from the solenoid (the wording in the answer is really bad, there is absolutely no such thing as "repel towards", only "repel away from").

Sorry for all my previous tedious explanations, I hope you understand this explanation of mine haha.

Best Regards
Happy Physics Land

Ahh, ok I see. Thanks for the explanation Happy Physics Land!

One last thing though: Why is the ring permanently repelled when the switch is closed. Wouldn't the ring only be momentary, similar to Faraday's experiment with the primary and secondary coil wrapped around the wooden block - because the changing magnetic flux is only changing when the power supply is turned on/off?

[jamonwindeyer]

Hey Meckenza!

Just a little clarification of HPL's answer.

He is correct in saying that flux flows to the left in the solenoid. However, don't think of it as there being an imaginary south pole on the inside of the solenoid.

Instead, consider it this way. A solenoid with a current flowing through acts as an electromagnet, and behaves very similarly to a bar magnet (especially for your purposes). For a bar magnet, the magnetic field lines form closed loops. That is, magnetic field lines (which represent the direction of magnetic flux) must always loop back on themselves. We also have to abide by the convention that field lines go out from North and go in at South. Logically, the only way for this to occur and the diagrams to work, is for the lines to loop outwards from North, around the magnet, into South, and back through to the beginning. There is no secret south pole involved, just the principle of closed magnetic field lines applied to a magnetic dipole.

As for your question on momentary vs ongoing repulsion. As the magnetic field turns on, the Eddy Currents flow to cause the ring to be repelled from the solenoid, as HPL discussed above. Now, the magnetic field stays the same, we have a DC power supply so once it is on, the solenoid acts as a bar magnet. This would lead me to believe that eddy currents stop flowing and the ring falls back into place after a momentary repulsion. That your answers suggest otherwise is interesting, it is definitely not what I would say would be occurring... I'm not sure!