Answer the question
In order to leave comments, you need to log in
How to estimate the charge of a conductor moving in a constant magnetic field?
How to estimate the volumetric charge density of a conductor moving in a constant magnetic field perpendicular to the lines of force?
For simplicity, let's take a bar 1 m long and with a section of 1 cm by 1 cm. We will move in a magnetic field at such a speed as to create a voltage of 10 V at the ends of this conductor. Inside the conductor, the charges will be redistributed so that the Lorentz forces balance the Coulomb forces. But the Coulomb force has in its denominator the dielectric constant of the medium, which shows the degree of weakening of the electric forces in the medium, and for metals this indicator is considered to be infinity, therefore, free electrons are also considered free and their Coulomb interaction is not taken into account in a number of calculations. Would it be logical to conduct such a thought experiment? Imagine that the conductor instantly stops and that charge which both relaxes and reflects the charge accumulated before? Thus, from the short circuit current and the drift velocity of charge carriers, we can estimate the maximum charge density at different ends of the conductor.
If this logic is correct, then we try to count.
On the one hand, we have the current I=dq/dt=Ro*dx*S/dt=Ro*S*V
where Ro is the volumetric charge density
V is the charge drift velocity under the action of voltage
S is the cross section of the conductor
On the other hand I = U/ R \u003d U / (Rr * l / S)
Rr - resistivity
l - conductor length
S - its cross section
Equating two expressions to each other, we get Ro \u003d U / (Rr * l * V)
Take for aluminum the resistivity Rr \u003d 28 * 10^-9 Ohm*m/m^2 and charge drift velocity V = 0.01 m/s. We get Ro = 3.5 * 10^10 C/m^3 or 3.5 * 10^4 C/cm^3 - in my opinion it's too much. In fact, this is the maximum possible concentration of free electrons for aluminum.
Answer the question
In order to leave comments, you need to log in
Didn't find what you were looking for?
Ask your questionAsk a Question
731 491 924 answers to any question