homer's physics corner

[Homer]

i dont quite understand

[BasicAcid]

What don't you understand? Have you learnt about the right hand grip rule yet?
Do you use the Heinemann textbook?
If so, thoroughly read pages 341 - 342 and if you still don't understand, go back over chapters 9.1 - 9.3 carefully.

[Robert123]

I think a better rule to use here is instead the right-hand solenoid rule. It works similar to the normal right hand grip rule but your fingers wrap around in the direction of the current and your thumb points in the direction of the magnetic field.

However, I'm pretty sure that the diagram is wrong since magnetic fielD lines are meant to point towards the South Pole, not the north.

[Homer]

i find it a bit hard like where do i wrap my right hand? diagram is terrible. the way ive been taught works for all the other solenoids but this one i get the wrong answer

[Homer]

Suppose we have a circuit with a 12V battery. Four identical lamps each consuming 18W of power are connected in parrallel to each other. What would be the current flowing through the circuit? Why? Also what would be the total power consumption of the circuit and why?

Thanks

[lzxnl]

Suppose we have a circuit with a 12V battery. Four identical lamps each consuming 18W of power are connected in parrallel to each other. What would be the current flowing through the circuit? Why? Also what would be the total power consumption of the circuit and why?

Thanks

Current = P/V = 18W/12V = 1.5A
This is current through each globe. Parallel circuit => sum currents together to give 6A

As for the power, P = VI = 12V*6A = 72W.

[SocialRhubarb]

Just to clarify, the right hand grip rule and the right hand solenoid rule are the same thing.

The diagram is correct, because, while Robert is right in saying that usually field lines point to the south pole, the right hand grip rule (or right hand solenoid rule) refers to the external field, not the internal field. Curl the fingers on your right hand in the direction that the current is flowing, and your thumb will point to the north pole of the external field generated by the solenoid.

If it helps, you can look at it from one end, and view the cross section of the solenoid as a circle. Then, you curl the fingers on your right hand in the direction of the current flow, and your thumb points towards the north pole of the external magnetic field.

[Homer]

An electron is accelerated from rest in a uniform electric field through a potential difference of 50.0 V.
What is the gain in kinetic energy of the electron?
The solution manual uses the formula

I am not sure HOW they got that.

[lzxnl]

An electron is accelerated from rest in a uniform electric field through a potential difference of 50.0 V.
What is the gain in kinetic energy of the electron?
The solution manual uses the formula

I am not sure HOW they got that.

Another failing of VCE physics; they don't even tell you what a potential difference is. Or, if they do, they don't tell you how to use them.

By definition, electric potential = electrical potential energy / charge
So change in potential energy = charge * change in electric potential
Neglecting sign issues, if we're only dealing with magnitudes, change in kinetic energy = charge * change in electric potential
or

[Homer]

Suppose we come across a question where a beam of light is sent through a narrow slit and projected on to a screen. The light is of a single wavelength (what does that mean?) and the wavelength of light is the same as the width of the slit.

I am not sure how the wavelength of the light and the width of the slit are related. I know the amount of diffraction equals width of the slit divided by the wavelength, however what does it mean to have a light of wavelength the same size as the width of the slit? Would there still be diffraction? Interference?  What if the slit was widened  by 10%, what effect would that have on the pattern?

[lzxnl]

Suppose we come across a question where a beam of light is sent through a narrow slit and projected on to a screen. The light is of a single wavelength (what does that mean?) and the wavelength of light is the same as the width of the slit.

I am not sure how the wavelength of the light and the width of the slit are related. I know the amount of diffraction equals width of the slit divided by the wavelength, however what does it mean to have a light of wavelength the same size as the width of the slit? Would there still be diffraction? Interference?  What if the slit was widened  by 10%, what effect would that have on the pattern?

Single wavelength => it can actually cause a diffraction pattern as opposed to white light which consists of a large number of wavelengths

This is yet another failing of VCE physics. They don't tell you much at all. It can be shown that for a single-slit diffraction pattern, the intensity minima occur at where d sin theta = m*wavelength where m is a non-zero integer.
Now, if d = wavelength, sin theta = m. If m is an integer, m can only be 1. I.e. light spreads out completely as there is no intensity minimum.
I think what VCE physics has done is look at the width of the first intensity minimum, i.e. where theta = inverse sine (wavelength/d)
You have it the wrong way around, Homer.
If you widened the slit by 10%, then your wavelength is smaller than the width. You will have an intensity minimum somewhere, so not "complete" diffraction.

[Homer]

so not "complete" diffraction.

would that mean the same pattern but the overall light intensity is less?

[lzxnl]

would that mean the same pattern but the overall light intensity is less?

It means you won't have the same pattern. In complete diffraction, the sound spreads out everywhere. The amount of spread will determine the intensity and uniformity of the diffraction pattern. If you have incomplete diffraction, then the intensity will drop to zero at the minimum.
You can hardly speak of "overall light intensity"...intensity of what? The pattern? The intensity isn't constant; it's a function of the slit width, the wavelength and the distance from the central maximum.

[Homer]

i was just wondering,

when we are standing on a building which in 100m high, the gravitational potential energy is 100mg (since mgh, where h is 100)
on ground we have 0 since h=0 but what if we are underground? :O where h is negative?

[Stevensmay]

h is the distance between the thing you will be impacting with and your actual height. So if I'm at the bottom of a hole, the distance is 0 thus I have no potential energy. Same with standing on the ground.

If I was 50m above ground level, and there happened to be a 50m hole underneath me then my kinetic potential is 100mg, as the effective height is 100m.

EDIT: See SocialRhubarb's post for a more in-depth explanation.
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