Compass in a solenoid, oscillating

[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?