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sergiokapone

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## 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.

## 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

$$\vec B = \frac{2I}{cr^2}(-y\hat e_x + x\hat e_y )$$

## 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:

$$U = 2\frac{e}{m} \frac{I^2}{c^2}\ln a/b$$

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