Compass in a solenoid, oscillating

[Vuldoraq]

Thats right. m isn't the mass by the way-it's still the magnetic dipole moment of the magnet (should call it $$m_{d}$$, which we will need to assume is unaffected by the B-field of the solenoid.

 

[TFM]

Okay so for I = 1 Amp, the period is 2 seconds

For I = 2 amps, the period is infinite

$$T = \sqrt{I \frac{4 \pi^2}{mB}}$$

$$B = \mu_0 n I_{current}$$

thus:

$$T = \sqrt{I \frac{4 \pi^2}{m\mu_0 n I_{current}}}$$

I: 1

$$2 = \sqrt{I \frac{4 \pi^2}{m \mu_0 n 1}}$$

$$4 = I \frac{4 \pi^2}{m \mu_0 n 1}$$

I:2

$$\infty = \sqrt{I \frac{4 \pi^2}{m\mu_0 n 2}}$$

Does this look okay?

TFM

 

[Vuldoraq]

Physically speaking it looks fine, but that infinity won't be much help to you.

What would the B field inside the solenoid be if the period of the magnet is infinite? (Newtons first law mighst help).

 

[TFM]

If the period is infinite, would that imply that the magnet is not rotating?

Newtons First Law:

A body will remain wuth constant motion unless other forces act upon it.

The forces are in balanced, but they are with all the rest as well.

TFM

 

[Vuldoraq]

Thats right-since the magnet is not accelerating the forces on it must be balanced. So the solenoids B-field (which is uniform inside) must equal the Earths B-field (which we can assume is uniform). From this and the formula for B-field inside a solenoid you can find an expression for $$n*\mu_{0}$$, which you will need to find the B-field inside the solenoid when the current is 1A.

 

[TFM]

So at 2 Amps, the B field = the Earth's Magnetic Field

B for the Earth is between 30 - 60 microteslars I'll use 30, since 60 is for near the poles.

$$B = \mu_0 nI$$

$$0.003 = \mu_0 n 2$$

$$n = \frac{0.003}{2\mu_o}$$

gives me the number of coils to be:

1193.7

seems a large number of coils, even for a solenoid?

TFM

 

[Vuldoraq]

Earth's B-field is half a gauss on average or 5*10^-5 T

 

[TFM]

that gives a far more reasonable answer of 19.89 (or 20) Coils. Whats the 30-60 MicroTeslar then?

Also, more importantly, for the equation:

$$T = \sqrt{I \frac{4 \pi^2}{m\mu_0 n I_{current}}}$$

we now know n!

so that just leaves Inertia and magnetic dipole moment.

I: 1 Amp

$$2 = \sqrt{I \frac{4 \pi^2}{m\mu_0 20 * 1}}$$

$$4 = \frac{I 4 \pi^2}{m\mu_0 20}}$$

$$80 \mu_0 = \frac{I 4 \pi^2}{m}}$$

$$\frac{80 \mu_0}{4 pi^2} = \frac{I }{m}}$$

Is this at all useful?

TFM

 

[Vuldoraq]

There is slight mistake, in your equation for T. You should have a value for the total B field felt by the magnet, rather than just the B-field due to the solenoid. B totla will be the vector sum of Earths b-field and the solenoids B-field, which we know are in opppisite directions.

 

[Vuldoraq]

that gives a far more reasonable answer of 19.89 (or 20) Coils. Whats the 30-60 MicroTeslar then?

TFM

I think you forgot that a micro is$$1*10^-6$$, 30-60 micro tesla is the Earths min and max B-field, as you said. 50 micro tesla is the average field across the whole planet (apparently)

 

[TFM]

Okay so:

$$T = \sqrt{I \frac{4 \pi^2}{m\mu_0 n I_{current}}}$$

should be:

$$T = \sqrt{I \frac{4 \pi^2}{m((\mu_0 n I_{current}) - (B_Earth))}}$$

?

TFM

 

[Vuldoraq]

Thats correct, I'm sorry I should have spotted it earlier. I would take the Earths field direction as positive though. You can see this makes sense because if you let the solenoid B = Earth's B you get a zero in the denominator which will make the period infinity, so our equation is consitent with what we know already.

 

[TFM]

Is this:

$$\frac{80 \mu_0}{4 pi^2} = \frac{I }{m}}$$

from the previous postings still correct/useful?

TFM

 

[Vuldoraq]

Is this:

$$\frac{80 \mu_0}{4 pi^2} = \frac{I }{m}}$$

from the previous postings still correct/useful?

TFM

Alas this is incorrect, although you still want to find and expression for I/m. Your value for n is still right though.

 

[TFM]

Okay so going back:

$$T = \sqrt{I \frac{4 \pi^2}{m((\mu_0 n I_{current}) - (B_Earth))}}$$

I: 1 Amp, T = 2 sec

$$2 = \sqrt{I \frac{4 \pi^2}{m((\mu_0 20 *1) - (50*10^{-5}))}}$$

$$4 = \frac{I4 \pi^2}{m((\mu_0 20 *1) - (50*10^{-5}))}$$

$$4((20\mu_0) - (50*10^{-5})) = \frac{I4 \pi^2}{m}$$

$$\frac{(80\mu_0) - (200*10^{-5}))}{4 \pi^2} = \frac{I}{m}$$

Does this look better now?

TFM

 

[Vuldoraq]

Almost, you seem to have gained an extra factor of ten in your value for Earth's B-field magnitude

$$2 = \sqrt{I \frac{4 \pi^2}{m((\mu_0 20 *1) - (50*10^{-5}))}}$$

(should be $$5*10^-5$$). If you make this correction then you'll get a value for I/m, which you can sub back into your time equation. You know B-solenoid is zero and B-earth=5*10^-5, so you should get an answer.

 

[TFM]

So close...

So:

$$2 = \sqrt{I \frac{4 \pi^2}{m((\mu_0 20 *1) - (5*10^{-5}))}}$$


$$4 = \frac{4 \pi^2 I}{m((\mu_0 20 *1) - (5*10^{-5}))}}$$


$$((\mu_0 80 *1) - (2*10^{-4}) = \frac{4 \pi^2 I}{m}$$


$$\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} = \frac{I}{m}$$

Does this look better?

TFM

 

[Vuldoraq]

Mooch better.

 

[TFM]

So now:

$$T = \sqrt{I \frac{4 \pi^2}{m((\mu_0 n I_{current}) - (B_Earth))}}$$

modify slightly:

$$T = \sqrt{\frac{I}{m} \frac{4 \pi^2}{((\mu_0 n I_{current}) - (B_Earth))}}$$

and then enter value of current, I = 0, and this should give me the period?

TFM

 

[Vuldoraq]

Hopefully, provided we've made the right assumptions and haven't missed anything. I wish someone else would check...

You'll need to put your value for I/m in there as well, don't forget.

 

[TFM]

Okay so:

$$T = \sqrt{\frac{I}{m} \frac{4 \pi^2}{((\mu_0 n I_{current}) - (B_Earth))}}$$

$$\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} = \frac{I}{m}$$

n = 20

Thus:

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 I_{current}) - (B_Earth))}}$$

Insert I = 0 Amp

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 * 0) - (B_Earth))}}$$

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 * 0) - (B_Earth))}}$$

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{-B_Earth}}$$

B earth: 5 x 10^-5

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} (-7.9 * 10^5)}$$

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} (-7.9 * 10^5)}$$

\mu_0 = 4pi * 10 ^-7

$$T = \sqrt{\frac{((80(4\pi * 10^{-7})) - (2*10^{-4})}{4 \pi^2} (-7.9 * 10^5)}$$

$$T = \sqrt{(-39.1) (-7.9 * 10^5)}$$

and:

$$T = \sqrt{3.08 * 10^7}$$

and:

T = 5553

Seems a bit big?

TFM

 

[Vuldoraq]

Okay so:

$$T = \sqrt{\frac{I}{m} \frac{4 \pi^2}{((\mu_0 n I_{current}) - (B_Earth))}}$$

$$\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} = \frac{I}{m}$$

n = 20

Thus:

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 I_{current}) - (B_Earth))}}$$

TFM

Here you have forgotten that the 4*pi^2 on the left cancels with the one on the right ( the I/m multiplies everything in the equation). Remember if you get wild results always check your arithmetic for mistakes, you can save a lot of marks on exams if always check (I often make silly mistakes).

 

[TFM]

Okay so:

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{1} \frac{1}{((\mu_0 20 I_{current}) - (B_Earth))}}$$

we can put this together:

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{((\mu_0 20 I_{current}) - (B_Earth))}}$$

I = 0

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{((\mu_0 20 * 0) - (B_Earth))}}$$

Thus

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{-B_Earth}}$$

insert mu_0

$$T = \sqrt{\frac{((80*4\pi *10^{-7}) - (2*10^{-4})}{-B_Earth}}$$

Insert B Earth:

$$T = \sqrt{\frac{((80*4\pi *10^{-7}) - (2*10^{-4})}{-5 * 10^{-7}}}$$

gives:

$$T = \sqrt{198}$$

T = 14.1


Seem better?

TFM

 

[Vuldoraq]

Okay so:

$$T = \sqrt{\frac{I}{m} \frac{4 \pi^2}{((\mu_0 n I_{current}) - (B_Earth))}}$$

$$\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} = \frac{I}{m}$$

n = 20

Thus:

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 I_{current}) - (B_Earth))}}$$

Insert I = 0 Amp

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 * 0) - (B_Earth))}}$$

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 * 0) - (B_Earth))}}$$

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{-B_Earth}}$$

B earth: 5 x 10^-5

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} (-7.9 * 10^5)}$$

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} (-7.9 * 10^5)}$$

\mu_0 = 4pi * 10 ^-7

$$T = \sqrt{\frac{((80(4\pi * 10^{-7})) - (2*10^{-4})}{4 \pi^2} (-7.9 * 10^5)}$$

$$T = \sqrt{(-39.1) (-7.9 * 10^5)}$$


TFM

Actually I think the mistake is here, my bad. This

$$T = \sqrt{\frac{((80(4\pi * 10^{-7}))$$ doesn't equal -39

 

[Vuldoraq]

Damn Latex! Your value for Earth's B-field is too small, should be 5*10^-5. Apart from that it looks good.

 

[TFM]

Okay so this should be:

$$T = \sqrt{\frac{((80*4\pi *10^{-7}) - (2*10^{-4})}{-5 * 10^{-5}}}$$

So this gives me:

$$T = \sqrt{1.989}$$

T = 1.41

were going from one extreme to the other here...

Does this look okay now?

TFM

 

[Vuldoraq]

Yeah it seems much better. It's less than the period when 1 amp flows, which it should be considering we said the solenoid was opposing the Earth's magnetic field. A good test would be too sub in a higher value of current, say 3 Amps and see what period you get for the oscillations.

 

[TFM]

Okay so testing it out:

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{4 \pi^2} \frac{4 \pi^2}{((\mu_0 20 I_{current}) - (B_Earth))}}$$

$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{((\mu_0 20 I_{current}) - (B_Earth))}}$$


I= 3:


$$T = \sqrt{\frac{((80\mu_0) - (2*10^{-4})}{((\mu_0 20 * 3) - 5 * 10^-5)}}$$

\mu_0 = 4pi * 10 ^{-7}

$$T = \sqrt{\frac{((80(4\pi * 10 ^{-7})) - (2*10^{-4})}{((4\pi * 10 ^{-7} * 20 * 3) - 5 * 10^-5)}}$$


$$T = \sqrt{\frac{-9.95 * 10^{-5}}{((4\pi * 10 ^{-7} * 20 * 3) - 5 * 10^-5)}}$$

$$T = \sqrt{\frac{-9.95 * 10^{-5}}{2.54 * 10^{-5}}$$

I've obviously done something wrong here, since it's giving me the square root of a negative?

TFM

 

[Vuldoraq]

No you did the right thing.

It could be our model isn't valid beyond 2 Amps, maybe we did something wrong or you can perhaps ignore the negative, since it pertains to the direction of the total B-field and we only want the magnitude. If you do ignore the negative you get 1.98 s, which is close to the period with 1 Amp, which you would expect it to be.

I'm not sure...

 

[Vuldoraq]

I think that when the solenoids field exceeds the Eaths the magnet will flip around and align with the solenoid instead. That means in our predefined coordinate system the magnet dipole gets reversed, and so another minus sign comes in and cancels the one from the solenoid field exceeding the Earths. Does that make sense?