Breaking Filaments using magnets.

[TFM]

Homework Statement

Bringing a strong magnet close to an operating incandescent light bulb has the effect of breaking the filament. Breakage might be expected when the extra force on the filament becomes comparable with gravity. Estimate the field strength likely to cause problems with a 60W bulb, in which the diameter of the tungsten filament is 0.1mm. (Density of tungsten = 19.3 x 10^3 kg.m^-^3.)

Homework Equations

Density = \frac{mass}{volume}

VI = P

\int B\dot dl = \mu_0 I_{encl} <--not sure on the relevance

F = q[E + v \times B] I think this should be equal to F = mg

The Attempt at a Solution

I can work out individual components, such as the current flowing through the wire (assuming that the source is 240V), but I can't seem to be able to make a connection between everything to get a useful formula.

Anyone got any ideas what would be the best thing to do first in this exercise?

TFM

 

[ak1948]

you use 240V light bulbs? The standard in the US is 120V.

Anyway you can work out the current. So figure the force in terms of the current rather than qv in your fourth equation. (Really force per unit length of the filament so you can equate it to weight per unit length).

 

[TFM]

Well here in England we have 240V mains, so I'm assuming the lights run off the same.

Anyway, the current is thus:

I = \frac{P}{V} = \frac{60}{240} = 0.25 Amps

so now I have to change the force Equation into Current:

F = q(E + v \times B) = qE +qV \times qB = qE + I \times qB

But I don't think this is quite right?

TFM

 

[ak1948]

I didn't mean to suggest merely substituting I for qv; rather that you find the appropriate equation using I and B. This must be in your text, its fundamental. Incidentally you'll be looking for the force per unit length, as well as the weight per unit length of the filament.

 

[TFM]

I can't seem to find it at the moment, but I have this equation in the back of my mind of

F = BIl, where B is the magnetic field, I the current and l the length of wire (Although this bit is unknown)

TFM

 

[ak1948]

Thats right. (Not precisely correct as a vector equation if that's what the bold indicates, but since you assume the field is at right angle to the current, its right using amplitudes).

So just find the weight for an equal length and equate.

 

[TFM]

I'm not using bold text, its just how Latex is displaying it (I use vector arrows above when diesgnating vectors, it makes it more clear). So

F = BIl

and

F = mg

I = 0.25 Amps

m = \pi * r^2 * l * \rho

Thus:

B = \frac{Il}{mg}

Insert m:

B = \frac{0.25 l}{\pi * r^2 * l * \rho g}

cancel the l

B = \frac{0.25}{\pi * r^2 * \rho g}

and inserting values gives:


B = 168

look about right?

Also, is the unit the weber?

TFM

 

[ak1948]

B is weber/m^2

 

[TFM]

Excellent, so the value for B is 168 Weber/m^2

Thanks,

TFM

 

[Vuldoraq]

I'm not using bold text, its just how Latex is displaying it (I use vector arrows above when diesgnating vectors, it makes it more clear). So

F = BIl

and

F = mg

I = 0.25 Amps

m = \pi * r^2 * l * \rho

Whoa hold your horses, up to here everything is fine. There is an algebra error in the equation below, don't forget your solving for B

Thus:

B = \frac{Il}{mg}

 

[TFM]

Thats a very silly mistake...It should be

mg = BIl

B = \frac{mg}{Il}

Thus:

B = \frac{\pi * r^2 * \rho g}{I}

giving a value of: 0.00428 Amps.

Does this look better?

TFM

 

[Vuldoraq]

Much better

However can you show me the exact numbers you used to get your numerical answer? Ie did you round Pi or anything? The reason I ask is because I got value of about 6*10^-3.

 

[TFM]

I used a spreadsheet to determine the value, so pi is as accurate as whatever accuracy ios given by it.

I used:

\frac{\pi * 0.00005^2 * (19.3 * 10^3) * 9.8}{0.25}

this gives me a value for the B field of 5.94 * 10^{-3}

(not the previous answer, I wrote the density the wrong way round )

TFM