# Rod ontop two parallel influenced by B field

Hey,

I've just been studying for my physics exam in a few days time and I'm trying to work out this problem,

a)the proof was straight forward and I wrote that the field is uniform over the cylinder because of cylindrical symmetry and c) I think was ok F = q(vxB) so q|v||B| going up the page so say z direction.

I'm just a bit stuck on the rest,

for b) I cant really picture how this think is set up, the field is into the page so by right hand rule the current flows to the right. But i'm not sure what its supposed to look like, would that mean the current is flowing anti-clockwise when you look at the cylinder from above? and I would just be found using the equation given, J = I/A = Eσ, we would probably need to find E, so using gausses law which a guassian cylinder inside the conductor, but then you would get the field at some radius within the cylinder involving a q which hasnt been provided in the question, like they didnt give a volume charge density. So i'm a little confused there.

For d) im really not sure where to start with it, could anyone point me in the right direction?

and for e) I would of said that the kinetic energy went to electric potential energy cause the system is changing so the currents would form to counteract the change and eventually want to pull the cylinder in the opposite direction

Heres the question 20/100 marks

http://img265.imageshack.us/img265/5363/32063619.jpg [Broken]

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Hi there!
It seems to me, from my prespective, that you had done everything properly, and as for your queries:
For C, it's best that you think of the force as acting F=ILB, on a current carrying wire.
And as for B, you must recall that the induced current is such that would have to negate the effects of the external field, in other words, as you said counter-clockwise.
In order to find the "elusive" E, remember that:
$\Phi = \vec{B}\cdot\vec{S}$
And that $E = -\frac{\partial \Phi}{\partial t}$
D would entail a differential equation, for your force, meaning: m*v'(t) = F, and in this case, you need to find said force as a function of the local, temporal, velocity.
e), remember that the wire has resistance as well.
Lastly, you might consider, for a)'s sake, Maxwell's relation:
$\vec{\nabla}\times\vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0$ meaning that E is uniform(i.e irrotational).
I hope that helps,
Daniel

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Thanks heaps Daniel,

doesn't the EMF,

ε=-$\frac{d\Phi}{dt}$=$\oint{E}\bullet{dl}$

so if I can find the flux and then take the derivative it = EMF so I can find E?

When i find the flux what would I use for the area, because the pole is sitting over the field right, so would it be the diameter of the cylinder times its length, because the field lines would only pass through the surface facing the field. Is that right?

so phi = BDl

but if i do the derivative of that wrt time phi = Dl dB/dt and is dB/dt zero because its a uniform field?

Hey,
For the purpose of the flux, you'll have to consider the area swept by the rod.
So imagine the horizontal axis as x(t), and thus the area carried by the rod is x(t)*l(the rectangle formed).
The flux is therefore B*x(t)*l.
Now consider that when deriving d(phi)/dt, x'(t) = v(t), and that's it.
Daniel

Alright so for b)

ε=-$\frac{d\Phi}{dt}$=-lv(t)B

and ε=IR

so I = -$\frac{lv(t)B}{R}$=-Aσv(t)B

does that seem right? i checked the units and got amps

for d, im struggling to get the exponential factor that was given

F = m v'(t)

qBv(t)= m v'(t)

$\frac{qBv(t)}{m}$ = v'(t) <-- so that looks right for an exponential

then m = ρmV

$\frac{qBv(t)}{ρV}$ = v'(t)

and ive spent ages trying to turn that into B^2σ/ρ

the closest i have gotten is B ρcharge densitym

i checked the units for this and they all cancel in the exponential aswell, so it might be correct,

i just cant find a way to turn Bρ into B^2σ

There must be some easy step that i just havent even thought of

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Hi.
First of all, as I've already said, the magnetic force you should've taken is F=I*L*B;
So, you're right in stating that the EMF,
$\epsilon = -\frac{d}{dt}\vec{B}\cdot\vec{S}=-Bvl=IR$
Next invoke the force, which is, as stated above I*L*B, to get:
$\rho_m V\dot{v}=I*l*B = -Bl\frac{Bvl}{R}$
Now substitute R from a) R = l/(sigma*A)
$\rho_m V\dot{v} = -Bl\cdot(BvA\sigma_c)$
A*l = Volume, therefore:
$\rho_m \dot{v} = -B^2v\sigma_c$
And that's all!
Let me know if you had trouble following these steps,
Thanks,
Daniel

Thanks heaps for all your help Daniel,

Sorry I didn't do it useing I lxB, I tried it once but I didnt get that far because I didn't sub in the I from b) cause im silly haha.

Thanks alot =]