[Bobboau]

ok, this question is a bit diferent than most in that I don't have the answer supplied, so I was hopeing to get some coroborative reasoning.
there is a coil that has a radius of 10cm (.1m), has a magnetic feild of .01T(3M N/C) at it's center, and the total crossectional area of all windings of the copper wire must not exeed .25cm^2, and use the minimum amount of power posable, how many turns of how thick wire should be used?

show why power is fixed and the previus question is irrelivent.

assumeing this coil is to be atached to a 12V car battery, what would be the best number of turns and why?




another question, this one is easier.

sence a magnetic feild cannot do any work on a free particle how can it do work on a wire carrying a current (like in a motor).

[Bobboau]

uh... guys?

[Col. Fishguts]

I'm not sure what you want to know exactly in the first one, but I know I have the formulas for that kind of stuff written down somewhere. I feel lazy right now .... to lazy to look it up.

EDIT: After reading through it a few times, I think I know what's asked. I'm still too lazy to look up the formulas, but since you're doing homework, I figure you should have them nearby. My thoughts:

- You have a formula relating the resulting field inside the coil to coil diameter, number of windings and current running through the coil. Note that the field is independant of wire diameter.

- The consumed power is P = R*I^2.  R depends on the wire length and diameter and specific resistance of the metal.

- So now you have an optimisation problem and I guess the resutling power will always be the same. Since to get the required field strength you can either have less coil windings and a higher current, or more windings and a lower current. But the power needed for the given field strength in a coil of given diameter should should always be the same (as long a s the coil is not very long)

I hope you can carry on from here.

As for the second one:
A magnetic field doesn't any work on a free charged particle because the resulting force is always perpendicular on the particle's speed vector. But in a wire carrying a current, the electrons cannot move freely, since they're bound to the wire.
They're travelling through the wire due to the voltage applied to them. When moving through the magnetic field, they experience a perpendicular force due to the magentic field. But since they cannot leave the wire (unless the magnetic field would be immensly powerful) a force on the wire is the result (they pull the wire with them, so to speak). The wire probabl wasn't moving at the beginning, so work is done on the wire.

[IPAndrews]

Apples fall out of trees. The end.

[Dark RevenantX]

Give me two years to get from Hon. Biology to Hon. Physics in my Junior year and i'll help you.

[Freespace Freak]

Give me two years to get from Hon. Biology to Hon. Physics in my Junior year and i'll help you.

Your in high school?  I didn't think there were people that young here on a game that's seven plus years old.

[Turambar]

OK guys, chew on this one.

A hydrogen atom at rest emits a Lyman alpha radiation.

a) compute the recoil kinetic energy of the atom

b) What fraction of the excitation energy of the n=2 state is carried by the recoiling atom?  (hint: use conservation of momentum)

[Herra Tohtori]

Well, just calculate the momentum of a photon with characteristic Lyman alpha wavelength. The hydrogen atom will have the same momentum. Based on the momentum, you can calculate the recoil velocity of the hydrogen atom. From recoil velocity you can count the kinetic energy of the hydrogen atom.

As to part b, the whole energy of the N2->N1 transition is used to kinetic energy of the atom (Ek) and energy of the photon emitted (Ef). Hence, the whole energy E=Ek+Ef.

Therefore, the fraction of the energy carried by recoiling atom is

Ek / E = Ek / (Ek + Ef).

Ef is easy to calculate, and kinetic energy of the atom was already calculated in part a.

[Turambar]

yeah, that one sounded harder than it looked, time for the next one.

a beam of 10-MeV protons is incident on a thin Aluminum foil 10^-6m thick. find the fraction of particles that are scattered through angles greater than  a) 10 degrees   b) 90 degrees


yeah, im only doing this cause im bored and have PY homework.

[Herra Tohtori]

Well, that's a phenomenon called Rutherford scattering or Coulomb scattering. I don't know really much about it but the name and the basic principle - electric forces affecting in the aluminium foil cause the passing protons to change direction.

Basically, it all boils down to probabilities.

If the proton does NOT hit an aluminium nucleus, it only deflects slightly due to electric forces between electron and proton.

If the proton hits a nucleus, it bounces off-direction at high angles, likely over 90 degrees, because of much greater mass of the nucleus and much greater force between positive nucleus and positive proton.

Now, just calculate the probability of hitting a nucleus, and you have the fraction of particles that are likely deflected more than 90 degrees.


Surprisingly, google search for "aluminium atom radius" brought up this page, which specifically handles rutherford scattering of aluminium. It uses Alpha particles as scattering particles, but it also gives an equation that can definitely be used with a proton as well, just change the mass, radius and charge to correspond with those of proton...




I don't have exact solutions for that, but I guess that would be the correct way to handle the problem.



Now, something interesting from mechanics.


How much power does a hovering helicopter with mass M and rotor blade length of R consume in normal temperature and pressure, in one g environment?

This one is on the series "foolingly easy-looking problems"...

[Turambar]

but i dont know anything about lift, or the rotor blade configuration, is this some supremely efficient rotor we're using to simplify the problem?

and there's the standard temperature and pressure, so we'd need to find the amount and type of air there for the rotor blades to be exerting a force on, and which are exerting the same force back on the rotor blades and it just goes into a complex mess.

[Flipside]

Well, it would need to be keeping a lift of 9.8m/s to be hovering, and blades use up a lot of energy in torque when you adjust rotor speed, but whilst the copter is hovering, the energy use should be pretty stable.

[Herra Tohtori]

Okay, let's simplify the problem a bit.

The situation is analogous to question "what is the power of rocket engine?"

While it may seem simple to find out what is the power of a rocket engine, it's not as simple as might seem.

So, what is the power of a rocket engine with following specs?

-Net thrust: F = 100 000 Newtons
-ejection velocity of propellant: v=1000 m/s

It's already possible to solve the problem from this information.

Tip: The definiton of force is F=dp/dt. The thrust is static, so you are allowed to use for example one second as dt, it doesn't change anything since the thrust is not variable in this problem.

Also, the rocket engine does W amount of work during the time t (for example a second); this work is converted to kinetic energy of the propellant flowing through the engine during the given unit of time.

[Dark RevenantX]

The calculus... it burns!  (i'l be burning badly for the next three years)

[Herra Tohtori]

You only need to get used to it.


Anyway, the energy consumption of a rocket engine can be calculated easily by defining the kinetic energy it gives to propellant per unit of time.

In our example, the rocket gives a velocity of 1000 m/s to mass M of propellant, and this produces a thrust of 100 000 Newtons.

First, we have do define the mass. This is done through

F = dp/dt

where dp = change of momentum. Also, dp = M dv, and now we have an equation

F = M dv/dt

We already know the velocity change (1000 m/s) and time change (one second can be used in this case), so we effectively have defined the mass flow through the engine per unit of time.

M/dt = F / dv

Substitution:

M/dt = 100 000 kgm/s^2 / 1000 m/s = 100 kg/s


Mass flow is thus 100 kg/s, and this mass is given speed of 1000 m/s.


Thus, the engine does W amount of work to give kinietic energy to defined amount of mass per second. So, work per second is produced kinitic energy of the propellant, per second.

Kinetic energy change of the propellant flowing through the engine in a unit of time (second) is

Ek= ½ M v^2 = ½*100 kg *1000^2 m^2/s^2 = 50*1 000 000  J = 50 000 000 Joules.

This amount of work is done in a second, thus P = 50 MW.

Thus, the power of our example rocket is whopping 50 MW.


It should also be noticed that when the mass flow is increased and velocity decreased, less power is needed to produce same net thrust... That is also the reason why gliders have longer wings than fighters.

A wing, or a helicopter rotor can ideally be seen as a thrust generating device similar to rocket engine - but they use their motion to displace M amount of air, when rocket engines use chemical reactions to displace M amount of propellant.

So, a longer wing can displace more air than a short wing. Thus, a short wing needs to give the displaced wing much more velocity downwards in order to achieve same lift as the long wing achieves with much less downwash velocity. Thus, the short wing uses much more energy to produce same amount of lift as the long wind - and this manifests as increased drag, more power is needed from the engines to produce the needed energy for lift.