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Movement of electron in an nonuniform E and B fields

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1. Homework Statement
The system consists of a long cylindrical anode of radius [itex]a[/itex] and a coaxial cylindrical cathode with radius [itex]b[/itex] ([itex]b <a[/itex]). On the axis of the system has a thread with a heating current [itex]I[/itex],
creates in the surrounding magnetic field.
Find the smallest potential difference between the cathode and anode, whereby thermoelectrons with zeros initial velocity will reach the anode.

2. Homework Equations
Equations of motion
$$\ddot{x} = \frac{e}{m}\frac{Ux}{r^2\ln{b/a}} - \frac{e}{mc}\dot{z} B_y$$
$$\ddot{y} = \frac{e}{m}\frac{Uy}{r^2\ln{b/a}} + \frac{e}{mc}\dot{z} B_x$$
$$\ddot{z} = \frac{e}{mc}(\dot{x} B_y - \dot{y} B_x)$$
and
$$\vec B = \frac{2I}{cr^2}(-y\hat e_x + x\hat e_y )$$

3. The Attempt at a Solution
From equation for [itex]\ddot z[/itex]I found
$$\dot z = \frac{2eI}{mc^2}\ln{\frac{r}{b}}$$
Near anode:
$$\dot z(r=a) = \frac{2eI}{mc^2}\ln{\frac{a}{b}}$$

From the Energy conservation law:
$$\frac12 m v_a^2 = eU$$

I get
$$U = \frac{2eI^2}{mc^4}\ln^2{a/b}$$

The solution in textbook is differ from above:
$$U = 2\frac{e}{m} \frac{I^2}{c^2}\ln a/b$$
 
Last edited:

haruspex

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Hi know nothing about the subject, but I can see that your answer, with c4 in the denominator, is dimensionally wrong. You should be able to find an error by going through your working to find where a dimensional inconsistency crept in.
 
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Dimention of charge from [itex]ma = q^2/r^2[/itex]
$$q^2 = \frac{g\cdot cm^3}{c^2}$$
Dim of U is [itex]q/cm[/itex]
Thus
$$\frac{q}{cm} = \frac{q\cdot q^2}{c^2\cdot g} \frac{c^4}{cm^4} = \frac{q\cdot \frac{g\cdot cm^3}{c^2}}{c^2\cdot g} \frac{c^4}{cm^4}.$$

It seems, with dimension everything ok.
 

haruspex

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Dimention of charge from [itex]ma = q^2/r^2[/itex]
$$q^2 = \frac{g\cdot cm^3}{c^2}$$
Dim of U is [itex]q/cm[/itex]
Thus
$$\frac{q}{cm} = \frac{q\cdot q^2}{c^2\cdot g} \frac{c^4}{cm^4} = \frac{q\cdot \frac{g\cdot cm^3}{c^2}}{c^2\cdot g} \frac{c^4}{cm^4}.$$

It seems, with dimension everything ok.
You've lost me completely with that analysis. Can you express all of these variables in the standard M, L, T, Q dimensional notation?
E.g. your ma=q2/r2 looks like it translates to MLT-2=Q2L-2, which is clearly not true.
What is g?
 

TSny

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I get
$$U = \frac{2eI^2}{mc^4}\ln^2{a/b}$$
This looks correct.
The units seem to check out (cgs system), although I don't follow your unit analysis in post #3.
 
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You've lost me completely with that analysis. Can you express all of these variables in the standard M, L, T, Q dimensional notation?
E.g. your ma=q2/r2 looks like it translates to MLT-2=Q2L-2, which is clearly not true.
What is g?
Ok. I use cgs units
Unit of square of charge
$$Q^2 = \frac{M\cdot L^3}{T^2}$$
Dimension of U in cgs is [itex]QL^{-1}[/itex]
Thus
$$\frac{Q}{L} = \frac{Q\cdot Q^2}{T^2\cdot M} \frac{T^4}{L^4} = \frac{Q\cdot \frac{M\cdot L^3}{T^2}}{T^2\cdot M} \frac{T^4}{L^4}.$$
 

haruspex

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Unit of square of charge
$$Q^2 = \frac{M\cdot L^3}{T^2}$$
Ok, I am guessing that this is a quantum analysis trick that fixes some physical constant which, to most physicists, has dimension, as a dimensionless 1, thereby deriving a translation between dimensions which would conventionally be taken as independent. E.g. c=1, so L=T, or something like that.
 

TSny

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Ok. I use cgs units
Unit of square of charge
$$Q^2 = \frac{M\cdot L^3}{T^2}$$
OK. One thing that confused me in post #3 was your use of ##c## to apparently denote seconds. You wrote ##q^2 = \frac{g \cdot cm^3}{c^2}##
 
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OK. One thing that confused me in post #3 was your use of ##c## to apparently denote seconds. You wrote ##q^2 = \frac{g \cdot cm^3}{c^2}##
Yes, sorry :sorry:. [itex]c[/itex] is the seсond, but in cyrillic notation, I forgot.
 

TSny

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