Conducting rod falling between vertical rods in magnetic field

[intervoxel]

Homework Statement

Two vertical conducting rods separated by a distance d are connected to a capacitor of capacitance C. Another horizontal conducting rod of mass m is released at position y=y0 at time t=0 always in contact with the two vertical ones. The set is immersed in a uniform magnetic field B perpendicular to both rods.

a) Calculate the acceleration; the velocity and the distance delta y at time t.
b) What is the induced current?
c) Analyse the balance of energy of the system.

Homework Equations

- standard electromagnetism formulas

The Attempt at a Solution

a)
- calculate the emf:

$$\varepsilon=\oint f_s \cdot dl$$
$$F_s=q(v\times B)$$
$$f_s=v\times B$$
$$\varepsilon=vBd$$

$$\varepsilon=V$$
Is the signal correct?

- calculate the acceleration:

$$a=g - F_M/m=g-[q(v\times B)]/m$$

Here I'm stuck: How can I get rid of q?

$$v=\int_0^t a dt$$

$$y=y_0+vt+(1/2)at^2$$

b)

$$I=dq/dt=CdV/dt=C\frac{d}{dt}(vBd)=CBd\frac{dv}{dt}=CBda$$

c)

- formulate energy balance:

$$mgy=(1/2)mv^2+(1/2)CV^2$$

 

[diazona]

The Attempt at a Solution

a)
- calculate the emf:

$$\varepsilon=\oint f_s \cdot dl$$
$$F_s=q(v\times B)$$
$$f_s=v\times B$$
$$\varepsilon=vBd$$

$$\varepsilon=V$$
Is the signal correct?

Signal?

If you're asking about $\varepsilon = vBd$, sure, that looks right. I would have used $\varepsilon = -\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}$ to get it, but the answer is the same either way.

- calculate the acceleration:

$$a=g - F_M/m=g-[q(v\times B)]/m$$

Here I'm stuck: How can I get rid of q?

Try $$\vec{F}_M = I\vec{L}\times\vec{B}$$ instead... at least, that's all I can think of. If that's what they're after, it seems a little strange that the problem asks you to calculate acceleration before induced current, unless they want you to leave $a$ in terms of $I$.

 

[intervoxel]

Oh, I see. More generally
$$\overrightarrow{F}=I\oint_C \overrightarrow{dl}\times \overrightarrow{B}$$
Thank you.