Magnetic pressure on a conductor

1. Jun 28, 2017

vishnu 73

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
i have done the first two parts but the third part confuses me
i am assuming the current is flowing in the z direction then current is parallel to the magnetic field then why would there be a force acting on it ?

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2. Jun 28, 2017

vishnu 73

3. Jun 28, 2017

vishnu 73

i am sorry for the poor quality
a constant 10MA current flows through the long hollow conducting cylinder initially of outer radius of 1cm and inner radius of a . A uniform external magnetic field of 1.0T is applied along the axis of the cylinder.for calculations you may assume the thickness of the cylinder is negligible
qn) calculate the force acting on a thin shell of current r and r + Δr where Δr << b-a ?

once again sorry

i am thinking there is no force as the current is parallel to the magnetic field

4. Jul 2, 2017

rude man

I don't understand why this statement is made. The thickness is b - a.
In the region a < r < b there is a second B field, one set up by the current, . As shown in your illustration, this field is in the θ direction. The ensuing force will be in the r direction and so will tend to expand the outer radius, which is why the problem states "initially of outer radius b".

You're right, the externally applied B field is in the z direction and so exerts no force on the cylinder.

5. Jul 3, 2017

vishnu 73

@rude man
thanks for replying it really means a lot to me and thanks for your interest
i have several doubts

1)
i am not really too familiar with cylindrical coordinates but if a current were to set up a magnetic field would it not be curling around it and not in the theta direction as shown in the diagram in this problem the theta is perpendicular to r unit vector and the wire ie magnetic field is originating from the wire which i thought is not possible am i missing something please help me clarify?

2)
this one confuses me too because i have been brought to believe that something is not affected by the fields it produces in this case the magnetic field does not affect the wire

6. Jul 3, 2017

rude man

"Curling around it' is in the theta direction! Just apply Ampere's law to a B loop within a < r < b.
That's a very general statement, and incorrect. It's true for example that a charge is not affected by its own E field alone but certainly a current can produce its own B field, including within the current that causes it! And so the Lorentz force can exert a force on the shell.

7. Jul 4, 2017

vishnu 73

sorry got confused by the diagram

thanks for the wonderful hint really helped me to see what is going on

now i know why it is because in this competition the use of integrals is not needed but can be used
with that being said i proceed by using integrals

by considering the cross sectional area of the cylinder
a ring of internal radius r and external radius of r + Δr
magnetic field in the ring
L - length of cylinder
J - current density
$B = \frac{J \pi(r^2 - a^2) \mu_o}{2 \pi r}\\ I = J 2 \pi r dr\\ df= I L B\\ df = L \frac{J \pi(r^2 - a^2) \mu_0}{2 \pi r}J 2 \pi r dr\\ df = L \mu_0 J^2 \pi(r^2 - a^2) dr$
is this the answer for the question
assuming it is correct (correct me if am wrong)

moving on to the next part of the question about calculating the pressure on curved walls

$\int df = \int _{r = a} ^b L \mu_0 J^2 \pi(r^2 - a^2) dr\\ F = L \mu_0 J^2 \pi \int _{r = a} ^b (r^2 - a^2) dr\\ F = L \mu_0 J^2 \pi [\frac{r^3}{3} - a^2 r]_a ^b\\ P = \frac{L \mu_0 J^2 \pi [\frac{r^3}{3} - a^2 r]_a ^b}{2 \pi b L}\\$
how am i supposed to get a muerical answer out of this help again?

edit:
or for pressure on curved surface am i only supposed to use the force on the outermost shell

8. Jul 4, 2017

rude man

the question is a bit nebulous IMO so I would say it's OK.
If you set up the coordinate system the way I described,
dI = dI k
so B =
B θ
and dF = I (dl x B)
which tells you the direction of dF.
Or you can pick a point. say x = r, y = 0 where B is in the +j direction so that k x j = -i
which come to think of it is radially inward and so will tend to contract the shell's radius rather than expand it as I told you earlier. Sorry!
This looks good.
Why can't you numerically evaluate your expression for P? You have all the numbers you need.
The force is exerted on all the current form a to b. At a the current is zero and at b it is max. When you integrated you got all the force on all the dr current elements.

9. Jul 5, 2017

vishnu 73

i dont have a

10. Jul 5, 2017

rude man

You're right, you don't, so leave the answer with a in it.

11. Jul 6, 2017

rude man

I was thinking that since the instructor said calculus was not needed, you could go as follows: the cylinder is very thin so the current I flows in a thinshell at r=b=1 cm. Then run your ampere's law contour just above r = b so the B field there is simply μ0I/2πb and the force would be F = IL x B directed into the cylinder, with the pressure p = F/2πLb. This isn't rigorous & may even be wrong but it would avoid calculus. Hard to decide what exactly the problem authors had in mind.

12. Jul 11, 2017

vishnu 73

i believe this is the total current through the conductor
a
and what is the current here how to calculate that
and why is it inside all along i have assuming the force was acting outwards

and sorry i am not really getting your method it would be very helpful if you could clarify it for me thanks

13. Jul 11, 2017

rude man

Total current as given: 10 mA.
Pick any point along the contour r. Let's pick x=+r, y=0, z anywhere 0<z<L. Current in the +z direction so B is in the +θ direction which at that point is the +y direction. So at that point,
F = iL x B
L = L k
B = B j
k x j = -i
in other words, inward.
You can pick any other point along the contour, the result will always be the same: the force is inward. The effect is to squeeze the cylindrical shell inwards all around the circumference.

14. Jul 12, 2017

vishnu 73

ok now i understand the direction and calculation yielded 32 giga pa which is one of the options but i still dont get the method

the magnetic field is caused by all of the current i suppose
then in why is it that in the expression F = I L B
I is also the total current is it because the cylinder is thin and that it can just be approximated as a thin wire

15. Jul 19, 2017

vishnu 73

um sir is the answer correct
nextly
the next part of the question asks
the cylinder is flexible and allowed to move
explain what happens to the area magnetic field and magnetic flux within the cylinder as the cylinder moves(eg. contracts or expands ) as time passes

16. Jul 19, 2017

rude man

You can think of the cylindrical shell as comprising a large number of parallel wires. Each thin wire carries a small portion ΔI of the total current I and is affected by the magnetic field due to all the remaining thin wires. So for each wire there is a small force BL ΔI. Adding all the wires' forces gives you the total force.

17. Jul 19, 2017

rude man

Just apply Ampere's law to whatever the shell changes to. I would assume finite thickness (b - a) to visualize this better.

18. Aug 1, 2017

vishnu 73

that was the original integration method but in the new method we seem to calculate the magnetic field of all the small wires as a whole and calculate the effect of magnetic field on all of the wires that is once again like a thin wire being affected by its own magnetic field
the reason why it strikes me odd u asked me to take the magnetic field at r = just more than b ie at that since we are outside of the wire all of the current contributre to the magnetic field and then you asked me to use that to calculate the force on the wire which seems like treating the magnetic field as an external one which is what striked me odd
anyway this question seems odd to me also without integration

19. Aug 1, 2017

vishnu 73

sorry for the late reply have been busy with academic matter exams are coming hope you understand thanks

well the area just increases or decreases accordingly
the magnetic field through the hollow part of the cylinder remains the same while the flux would change accordingly as well
what is going on here this just seems to me wierd

the external magnetic field in the question doesn't seem to serve any purpose
why would increasing the area have any effect please explain

20. Aug 1, 2017

rude man

Well, as I said, the only good way to see this is to assume a finite thickness (b - a). Then the force pulling the cylinder inwards is zero at a and max. at b and yes to get the right answer you have to integrate this force from a to b.

Maybe they wanted you to take the mid-point of this force, I don't know. If you assume an arbitrarily thin wall I agree it's not obvious how a force is generated on the cylinder wall. I'm not sure where you'd put your contour to invoke Ampere's law. With a finite thickness it's obvious.