Generic magnetron IV curve As discovered by "Applied Science" YouTube channel (this video: https://youtu.be/I2k2g00onL0?t=470) typical magnetron found in household microwave ovens have quite steep IV curve after 4kV, with almost no current under 4kV.
1) What causes this steep rise of current after this threshold voltage? Is this some sort of "resonance"? Would it be possible to find other "resonance" voltages, which are lower or higher than 4kV?
2) What causes delay (few milliseconds) of current after voltage is applied? I can't believe electron lifetime in magnetron is in the order of milliseconds...
Best regards,
Mikhail. 
 A: 1)
No, the threshold voltage, known as the Hull cutoff voltage, is not a resonance effect, and there is only one such voltage for a given magnetron.  Instead, the Hull cutoff voltage is the voltage which is just sufficient to overcome the effect of the magnetic field, which tends to cause electrons which are emitted from the cathode to return to the cathode.

The above diagram represents a cross section of the magnetron, with the magnetron's axis pointing perpendicular to the plane of the diagram.  The inner circle, of radius $r_c$, represents the outer surface of the cathode, and the outer circle, of radius $r_a$, represents the inner surface of the anode.  I didn't draw the cavities in the anode, because the cavities are of little importance in a discussion of the cutoff voltage.
If the magnet is removed from a magnetron, electons emitted from the cathode can easily flow to the anode with just a small voltage, as in the path labelled $B=0$ in the diagram.  However, a magnetron normally includes a permanent magnet which produces a magnetic field along the axial direction, i.e., perpendicular to the plane of the diagram above.  From the form of the Lorentz force equation,
$$\mathbf{F} = q\mathbf{E} + q\mathbf{v} \times \mathbf{B}\ \ ,$$
the effect of a magnetic field $\mathbf{B}$ along the magnetron's axis will produce a force on an electron that's perpendicular to the axis, causing the electron to tend to curve around back towards the cathode.  If there's a sufficient voltage difference between the cathode and anode, the electron will still manage to reach the anode, as in the path labelled $V>V_c$, i.e., a current will flow.  But if the voltage isn't high enough, as in the path labelled $V<V_c$, the magnetic field will curve the electron's path so much that it will return to the cathode, i.e., no current will flow.
To calculate the cutoff voltage, it's convenient to use cylindrical coordinates, due to the cylindrical symmetry of the magnetron.  In cylindrical coordinates in general, the component of acceleration that's in the direction of the $\phi$ unit vector can be written
$$a_{\phi}=\frac{1}{r}\frac{d}{dt}\left(r^2 \frac{d\phi}{dt}\right)\ \ .$$
The component of the Lorentz force equation that points in the $\phi$ direction is thus
$$m \frac{1}{r}\frac{d}{dt}\left(r^2 \frac{d\phi}{dt}\right)=eB\frac{dr}{dt}\ \ ,$$
where $e$ is the elementary charge, $m$ is the electron's mass, and $B$ is the component of the magnetic field in the $z$ direction, which is the only nonzero component of $\mathbf{B}$. Positive $B$, i.e. $\mathbf{B}$ pointing up out of the diagram toward the viewer, produces counterclockwise electron rotation as shown the diagram.
The preceding equation can be rearranged as
$$\frac{d}{dt}\left(r^2 \frac{d\phi}{dt}\right)=\frac{eBr}{m}\frac{dr}{dt}=\frac{eB}{2m}\frac{d}{dt}r^2\ \ .$$
Integrating with respect to time gives
$$r^2 \frac{d\phi}{dt}=\frac{eBr^2}{2m}+c\ \ ,$$
where $c$ is an integration constant to be determined.  But right when the electron leaves the cathode, i.e., when $r=r_c$, the electron doesn't have a velocity in the $\phi$ direction yet, i.e. $d\phi/dt=0$ , which implies that
$$c=-\frac{eBr^{2}_{c}}{2m}\ \ ,$$
i.e.,
$$r^{2}\frac{d\phi}{dt}=\frac{eB}{2m}\left(r^2 -r_c^2\right)$$
or
$$\frac{d\phi}{dt}=\frac{eB}{2m}\left(1-\frac{r_c^2}{r^2}\right)\ \ .$$
Right at the cutoff voltage $V_c$, the path of the electron is such that when it reaches the anode, the velocity of the electron is essentially entirely in the $\phi$ direction, and has magnitude
$$v=r_a\frac{d\phi}{dt}\ \ .$$
Since the magnetic force on a electron is always perpendicular to the electron's velocity, the magnetic field performs no work on the electron.  Hence, the electron's kinetic energy when it hits the anode is due entirely to the work performed on the electron by the electric field.  At the cutoff voltage, this kinetic energy of the electron at the anode is
$$\frac{1}{2}mv^2=\frac{1}{2}m\left(r_a \frac{d\phi}{dt}\right)^2=eV_c\ \ .$$
Substituting in the above expression for $d\phi/dt$ at $r_a$ gives
$$\frac{1}{2}m\left[r_a \frac{eB}{2m}\left(1-\frac{r_c^2}{r_a^2}\right)\right]^2=eV_c\ \ .$$
Solving that equation for $V_c$ gives
$$V_c=\frac{eB^2 r_a^2}{8m}\left(1-\frac{r_c^2}{r_a^2}\right)^2\ \ .$$
2)
The creator of that youtube video says that he thinks that the 2ms lag between the voltage and the current might be an artifact of the current probe, that there might be some low-pass circuitry in the probe that's causing the delay.  That seems like a reasonable guess to me, particularly since there's not only a lag, but when the current in the plot does change, it changes "slowly" (over a couple ms) instead of changing sharply like the voltage plot does.
