A system consists of a long cylindrical anode of radius $a$ and a coaxial cylindrical cathode of rad

Question

A system consists of a long cylindrical anode of radius $a$ and a coaxial cylindrical cathode of radius $b$ ($b < a$). A filament located along the axis of the system carries a heating current $I$ producing a magnetic field in the surrounding space. Find the least potential difference $V$ between the cathode and anode at which the thermal electrons (mass $m$, charge magnitude $e$) leaving the cathode without initial velocity start reaching the anode.

Professor PC Thomas Classes · Answer

1. Coordinate System and Magnetic Field: Let the axis of the cylinders be the y-axis, and let x be the radial distance from the axis. The magnetic field produced by the heating current $I$ at a radial distance $x$ is azimuthal, with magnitude:
$B = \frac{\mu_0 I}{2\pi x}$
2. Force Analysis: As the electron moves outwards, the magnetic field exerts a Lorentz force. The component of this force in the axial (y) direction is:
$F_y = F_m \cos\alpha = evB \cos\alpha$
where $\alpha$ is the angle the velocity vector makes with the radial x-axis. From this geometry, $v \cos\alpha = v_x$, which is the radial velocity. Thus:
$F_y = e B v_x$
3. Equation of Motion: Apply Newton's Second Law in the y-direction:
$m \frac{dv_y}{dt} = e \left(\frac{\mu_0 I}{2\pi x}\right) v_x$
Using the chain rule, we can rewrite the acceleration as $\frac{dv_y}{dt} = \frac{dv_y}{dx} \frac{dx}{dt} = \frac{dv_y}{dx} v_x$. Substituting this into the equation:
$m \frac{dv_y}{dx} v_x = e \left(\frac{\mu_0 I}{2\pi x}\right) v_x$
Canceling the radial velocity $v_x$ from both sides gives:
$m \, dv_y = \frac{e \mu_0 I}{2\pi x} dx$
4. Integration: Integrate from the cathode ($x = b$) where the initial axial velocity is $0$, to the anode ($x = a$) where the final axial velocity is $v_0$:
$\int_0^{v_0} dv_y = \frac{e \mu_0 I}{2\pi m} \int_b^a \frac{dx}{x}$
$v_0 = \frac{e \mu_0 I}{2\pi m} \ln\left(\frac{a}{b}\right)$
5. Conservation of Energy: The potential difference $V$ does work on the electron, providing its kinetic energy. For the least potential difference required to just reach the anode, the radial velocity at the anode will drop to zero. Thus, its entire kinetic energy at the anode is solely due to its axial velocity $v_0$:
$eV = \frac{1}{2} m v_0^2 \implies v_0 = \sqrt{\frac{2eV}{m}}$
6. Solve for Potential Difference: Equating the two expressions for the axial velocity $v_0$ yields:
$\sqrt{\frac{2eV}{m}} = \frac{e \mu_0 I}{2\pi m} \ln\left(\frac{a}{b}\right)$
Squaring both sides and rearranging to solve for $V$:
$\frac{2eV}{m} = \frac{e^2 \mu_0^2 I^2}{4\pi^2 m^2} \ln^2\left(\frac{a}{b}\right)$
$V = \frac{e \mu_0^2 I^2}{8\pi^2 m} \ln^2\left(\frac{a}{b}\right)$