Okay, I have a few questions... (1 Viewer)

[Shadowless]

So I've recently completed all the multiple choice questions from the HSC Online Multiple Choice site and I have a few questions...

2001 Q08: [Complete]
Could someone explain? Thanks.

2002 Q10:
Could someone explain? Thanks.

2005 Q06: [Complete]
Could someone explain? Thanks.

2005 Q08: [Complete]
Could someone explain? Thanks.

2005 Q09: [Complete]
Why isn't it B?

2005 Q13: [Complete]
Could someone explain? Thanks.

2005 Q15: [Complete]
Why isn't it D?

[to be continued...]​

 

[cheepy5]

Do you have the sucess one book for physics, they explain the answers for multiple choice.

 

[Shadowless]

[continued...]​

2006 Q06: [Complete]
Could someone explain? Thanks.

2006 Q09:
Could someone explain? Thanks.

2006 Q10:
Could someone explain? Thanks.

2006 Q12:
Could someone explain? Thanks.

2007 Q08:
Could someone explain? Thanks.

2008 Q15:
Could someone explain? Thanks.

2009 Q03:
What exactly is the difference between C and D, and would I still be wrong in saying the answer's D?

2009 Q12:
Could someone explain? Thanks.

 

[Shadowless]

Do you have the sucess one book for physics, they explain the answers for multiple choice.

Nope. =(

 

[GoldyOrNugget]

This refers to your first post.

1. Connecting the ends allows a current to flow. By Lenz' law, this current is induced to oppose the motion that caused it i.e. to oppose the pendulum's swinging.

2. When the coil begins to rotate at P, it is cutting the maximum amount of flux lines, so at P the induced emf is a maximum by Faraday's law. Rotating at Q, it is cutting the minimum amount of flux lines (because the velocity of the coil at this time is parallel to the magnetic field) so there is 0 induced emf at this point. Only option B corresponds to this.

3. Length of rotating wire increases the amount of wire that is cutting magnetic flux. Speed of rotation affects rate of flux-cutting. Orientation affects whether the wire is parallel or perpendicular to field lines. Thickness is the only one that doesn't affect rate of magnetic flux cutting.

4. When the DC current is turned on, the secondary coil will experience an increase in magnetic flux. By Faraday's law, an emf will be induced. However, because there is no more changing flux, this emf will disperse quickly, resulting in the spike of option C.

5. By HSC logic, no eddy currents are induced in ring R because of the gap. In real life, option B would probably be correct.

6. I've never understood what these diagrams correspond to. The way I remember it is if the dopant level is closer to the valence band, then the dopant is associated with positiveness i.e. a p-type semiconductor. If it's closer to the conduction band, it's associated with things moving in the conduction band i.e. electrons i.e. an n-type semiconductor.

7. Electrons and holes move in OPPOSITE directions. Using the right hand rule (with the thumb in the positive direction), it can be seen that both electrons and holes will move downwards.

 

[cheepy5]

2001 Q08:
D because when you connect the two ends together an induced emf can flow through, due to lenz's law the magnetic effect of the induced emf will oppose the motion.

2002 Q10:
B, because when its parallel to the field it cuts through the magnetic flux at a max rate, therefore max induced current. At Q min ect therefore cos wave.


2005 Q6;
B, because changing the thickness does nothing to affect the induced emf, therefore nothing to the current.

2005 Q8:
C, its a Dc power supply so there is initial a change in flux when turned on .

 

[GoldyOrNugget]

Second post:

1. This is similar to Faraday's first motor. Consider only the radial magnetic field emanating from the north pole. So with the right-hand palm rule, your fingers should be pointing horizontally outwards toward the rod. The current moves down i.e. thumb is pointed downwards. It can be seen that at any point in time, the force on the wire by the motor effect is pushing it in a circle around the magnet.

2. Consider the Lorentz force (i.e. "a charged particle in a magnetic field experiences a force") on an electron on the disc, as that area of the disc is entering the field. Point your thumb in the opposite (positive) direction: out of the page. Orient your fingers with the magnetic field. Your palm will face upwards, meaning electrons move upwards, effectively meaning conventional current will move downwards through Y, through the external circuit and back to X.

3. The primary coil experiences a constant DC current. The secondary coil experiences short changes in magnetic flux when the DC current is switched on and off. This corresponds to option A.

4. By F=qvBsin(@), the velocity component in the direction parallel to the fields is not relevant. This is what the sin(@) is for. When it is travelling parallel to the fields, @=0 degrees therefore sin(@) = 0.

Someone else do the rest.

 

[Shadowless]

This refers to your first post.

1. Connecting the ends allows a current to flow. By Lenz' law, this current is induced to oppose the motion that caused it i.e. to oppose the pendulum's swinging.

2. When the coil begins to rotate at P, it is cutting the maximum amount of flux lines, so at P the induced emf is a maximum by Faraday's law. Rotating at Q, it is cutting the minimum amount of flux lines (because the velocity of the coil at this time is parallel to the magnetic field) so there is 0 induced emf at this point. Only option B corresponds to this.

3. Length of rotating wire increases the amount of wire that is cutting magnetic flux. Speed of rotation affects rate of flux-cutting. Orientation affects whether the wire is parallel or perpendicular to field lines. Thickness is the only one that doesn't affect rate of magnetic flux cutting.

4. When the DC current is turned on, the secondary coil will experience an increase in magnetic flux. By Faraday's law, an emf will be induced. However, because there is no more changing flux, this emf will disperse quickly, resulting in the spike of option C.

5. By HSC logic, no eddy currents are induced in ring R because of the gap. In real life, option B would probably be correct.

6. I've never understood what these diagrams correspond to. The way I remember it is if the dopant level is closer to the valence band, then the dopant is associated with positiveness i.e. a p-type semiconductor. If it's closer to the conduction band, it's associated with things moving in the conduction band i.e. electrons i.e. an n-type semiconductor.

7. Electrons and holes move in OPPOSITE directions. Using the right hand rule (with the thumb in the positive direction), it can be seen that both electrons and holes will move downwards.

I still don't really understand 2. Doesn't it cut the MAXIMUM amount of flux lines when the coil is PERPENDICULAR (i.e. at Q)?

But for the rest I get them now. Thanks!

 

[kiinto]

I still don't really understand 2. Doesn't it cut the MAXIMUM amount of flux lines when the coil is PERPENDICULAR (i.e. at Q)?

But for the rest I get them now. Thanks!

Maximum flux is when the coil is perpendicular to the magnetic field. But induced emf isn't about the amount of flux threading the coils, it's about the rate of change of flux through the coils. The rate of change of flux is at maximum where flux is at zero. (i.e. emf is equal to the negative gradient of the flux.)

 

[Vicky_x]

I am definitely not prepared for this exam...

 

[oh well]

I am definitely not prepared for this exam...

lol