Compass in a solenoid, oscillating

[TFM]

Homework Statement

A compass, consisting of a small bar magnet resting on a frictionless pivot through its centre, is placed in the middle of a long solenoid, which in turn is aligned with its axis pointing North-South. With a current of 1 Amp passing through the solenoid, a small displacement of the compass from its equilibrium orientation causes it to oscillate with a period of 2 seconds. If a current of 2 Amps is used, the restoring torque is zero and the period is infinite. Calculate the period of the oscillation when no current flows in the solenoid.

Homework Equations

Magnetic Field in a solenoid:

B=\mu_0 nI

Where B is the magnetic field, n the number of coils, and I the current


Magnetic Force:

F = q(E+v\times B)

Torque on a magnetic dipole:

G = m\times B

where G is torque

Torque:

torque = I\alpha

where I is the Moment of inertia

Oscillation:

x = A cos(\omega_0t + \phi)
v = A\omega sin(\omega_0t + \phi)
a = -A\omega^2 cos(\omega_0t + \phi)

\omega_0^2 = \frac{k}{m}

The Attempt at a Solution

The equation doesn't give you some of the required variables for the formulas I believe you need; most importantly, it doesn't give you the number of coils, so you can work out the B field.

So far, I have divided it into sections:

1 Amp:

B = \mu_0 n*1

G = m\times \left(\mu_0 n*1 \right)

2 Amp:

B = \mu_0 n*2

G = m\times \left(\mu_0 n*2 \right)

0 Amp:

B = \mu_0 n*0 = 0

There is no B field with no current

G = m\times \left(\mu_0 n*0 \right)

No torque from magnetic dipole

does this look at all right?

TFM

 

[TFM]

Does this look correct

?

TFM

 

[TFM]

What should i be doing then?

TFM

 

[Vuldoraq]

Greetings,

Your going in the right direction with this problem. I am not certain that the magnetic dipole will be zero though (don't forget the B-field of the magnet arises from all the tiny currents inside the magent, so there is a currrent and there should be a diplole). You need to apply Newtons second law using the torque due to a (magnetic) dipole and the usual expression for torque (I*alpha).

Once you've done this you should end up with a second order O.D.E. , which you can then solve by letting theta=something.

 

[TFM]

okay, so:

I \alpha = m \times B

B = \mu_0 n I

thus:

I \alpha = m \times \mu_0 n I

should I assume they are perpendicular to give:

I \alpha = m \mu_0 n I

also, the moment of inertia for a cuboid is:

I = \frac{m(b^2 + c^2)}{12}

but we aren't given the dimensions of the magnet

?

TFM

 

[Vuldoraq]

Sorry you can just leave the moment of inertia as I, you'll find an expression later that takes care of the unknowns. Edit: sorry they are are not perpendicualr here! You can use the small angle approxiamtion ie: sin(theta)=theta. Also keep B as B in the expression.

Now how are theta nd alpha realted?

 

[TFM]

Okay so:

I_m \alpha = m \mu_0 n I

So now I need to find a way to get the period from this equation.

Does this mean I need to put the angular version of this equation:

x = A cos(\omega_0t + \phi)

into the previous one?

TFM

 

[Vuldoraq]

Let me elaborate, you take the cross product as being equal to mBsin(theta) (ie just the magnitude of the cross product). This is a restoring torque so it's sign should be negative.

 

[TFM]

so it should be:

I_m \alpha = -m \mu_0 n I

?

TFM

 

[Vuldoraq]

It should be;

I \alpha=-mBsin(\theta)

=-mB\theta for small \theta

can you see why?

Note:I'm not expanding B to avoid confusion with inertia and also it's tidier ans we don't need to expand it yet.

 

[TFM]

I see,

I get that, its the small angle substitution, because in radians, the sin of small angles is near enough the same as the angle.

so:

I \alpha = -mB\theta

so now I think I have to get a differential in there to make it into a second order ODE, probably:

\alpha = ddot{\theta}

\alpha = \ddot{\theta} = \frac{d^2\theta}{dt^2}

Look right?

TFM

 

[Vuldoraq]

Thats right. Now you can solve this equation if you let \theta=\theta*sin(\omega*t)

 

[TFM]

So:

I \alpha = -mB\theta

and thus:

I \frac{d^2\theta}{dt^2} = -mB\theta

Thats right. Now you can solve this equation if you let \theta=\theta*sin(\omega*t)

I'm not sure, were did you get the expression from?

TFM

 

[Vuldoraq]

Well the compass is moving with simple harmonic motion (we assume), and so it's displacement \theta at any point will be related to it's angular frequency and the time at which we are looking at the situation. The equation is identical to the one in your first post, except the phase angle is zero and I'm saying the amplitude is just theta (which it should be in this case).

 

[TFM]

I see so:

\theta = \theta cos(\omega_0t + \phi)

Assume no starting phase:

\theta = \theta cos(\omega_0t)

I see now. I was confused slightly by the fact that you had two thetas equalling each other. Maybe I should call the Amplitude theta theta_0, as to not confuse myself:

\theta = \theta_0 cos(\omega_0t)

okay, so now:

I \frac{d^2\theta}{dt^2} = -mB\theta

I \frac{d^2\theta}{dt^2} = -mB(\theta_0 cos(\omega_0t))

now I have to rearrange the differential?

TFM

 

[Vuldoraq]

Yeah, I should have made that clearer, sorry. Now you solve like you would any O.D.E. in which you are using a substitution: just differentiate your expression for theta to find theta double dot in terms of this substitution and then sub all new expressions into the original equation (if that makes sense?).

 

[TFM]

so we now have to differentiate:\theta = \theta_0 cos(\omega_0t)

\theta = \theta_0 cos(\omega_0t)

Edit: the above should be theta, not d theta.

twice?

giving:\frac{d\theta}{dt} = t \theta_0 sin(\omega_0t)

and

\frac{d\theta}{dt} = -t^2 \theta_0 cos(\omega_0t)

?

TFM

 

[Vuldoraq]

Ooops you should be differentiating wrt to t rather than omega.

\frac{d\theta}{dt}=-\theta_{0}*\omega*sin(\omega*t)

\frac{d^{2}\theta}{dt^{2}}=-\theta_{0}*\omega^{2}cos(\omega*t)

 

[TFM]

I thought it should have been omega coming out, since that is what happens with the wave equation. so now do I insert this value of

\frac{d^2 \theta}{dt^2}

into the previous equation,

I \frac{d^2\theta}{dt^2} = -mB(\theta_0 cos(\omega_0t))

TFM

 

[Vuldoraq]

Thats right, then you should be able to cancel down and get a simpler expression.

 

[TFM]

okay, so we have:

I \frac{d^2\theta}{dt^2} = -mB(\theta_0 cos(\omega_0t))

\frac{d^{2}\theta}{dt^{2}}=-\theta_{0}*\omega^{2}cos(\omega*t)

Thus:

-I \theta_{0}*\omega^{2}cos(\omega*t) = -mB\theta_0 cos(\omega_0t)

Some parts will cancel:

-I \omega^{2}cos(\omega*t) = -mB cos(\omega_0t)

and:

-I \omega^{2} = -mB

Looking okay so far?

TFM

 

[Vuldoraq]

Looks good to me.

 

[Vuldoraq]

Looks good to me.

 

[TFM]

so what term do I need to rearrange for?

TFM

 

[Vuldoraq]

We're looking for the period so we want an expression that involves T.

 

[TFM]

Okay so:

-I \omega^{2} = -mB

and

\omega = \frac{\theta}{T}

so:

-I (\frac{\theta}{T})^{2} = -mB

-I \frac{\theta^2}{T^2} = -mB


I \frac{\theta^2}{mB} = T^2

giving:

T = \sqrt{I \frac{\theta^2}{mB}}

Does this look okay?

TFM

 

[Vuldoraq]

That theta should be 2*Pi, \omega=2*\pi /T

Apart from that your expression looks good, although I would be happier if another PF'er could confirm this, just to be on the safe side? (I am prone to making mistakes, so my apoligies).

 

[TFM]

Okay so:

-I \omega^{2} = -mB

and

\omega = \frac{2 \pi}{T}

so:

-I (\frac{2 \pi}{T})^{2} = -mB

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


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

giving:

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

and T is the period, so now, assuming all is well, do I stick in the values given at the beginning to find values for I and n, and then use these to find the period for when I = 0?

TFM

 

[Vuldoraq]

That's definitely the way to go.

 

[TFM]

Do we just need to find a ratio, because we don't know the mass m , or the number of coils n or the Moment of Inertia I

?

TFM

 

[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 is1*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

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

(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.