Electromotive Force Homework: Vertical Conducting Rods & Uniform 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:

<br /> \varepsilon=\oint f_s \cdot dl <br />
<br /> F_s=q(v\times B)<br />
<br /> f_s=v\times B<br />
<br /> \varepsilon=vBd <br />

<br /> \varepsilon=V <br />
Is the signal correct?

- calculate the acceleration:

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

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

<br /> v=\int_0^t a dt <br />

<br /> y=y_0+vt+(1/2)at^2 <br />

b)

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

c)

- formulate energy balance:

<br /> mgy=(1/2)mv^2+(1/2)CV^2 <br />

 

[diazona]

The Attempt at a Solution

a)
- calculate the emf:

<br /> \varepsilon=\oint f_s \cdot dl <br />
<br /> F_s=q(v\times B)<br />
<br /> f_s=v\times B<br />
<br /> \varepsilon=vBd <br />

<br /> \varepsilon=V <br />
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:

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

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
<br /> \overrightarrow{F}=I\oint_C \overrightarrow{dl}\times \overrightarrow{B}<br />
Thank you.