(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

- standard electromagnetism formulas

3. The attempt at a solution

a)

- calculate the emf:

[tex]

\varepsilon=\oint f_s \cdot dl

[/tex]

[tex]

F_s=q(v\times B)

[/tex]

[tex]

f_s=v\times B

[/tex]

[tex]

\varepsilon=vBd

[/tex]

[tex]

\varepsilon=V

[/tex]

Is the signal correct?

- calculate the acceleration:

[tex]

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

[/tex]

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

[tex]

v=\int_0^t a dt

[/tex]

[tex]

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

[/tex]

b)

[tex]

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

[/tex]

c)

- formulate energy balance:

[tex]

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

[/tex]

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# Homework Help: Electromotive force

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