Pho.rs: Hall Effect in InSb

A1 ^?? Determine what the magnitude and the direction of the electric field in the bar is, to yield the current described above.

First the electron velocity is calculated from the current $I$:
\[I = jS = ne_0 v bc, \quad v = \frac{I}{ne_0bc} = 25~\text{m}/\text{s}\] The components of the electric field are obtained from the electron veloc- ity. The component in the direction of the current is
\[E_\parallel = \frac{v}{\mu} = 3.2~\text{V}/\text{m}\] The component of the electric field in the direction $b$ is equal to the Lorentz force on the electron divided by its charge:
\[E_\perp = vB = 2.5~\text{V}/\text{m}\] The magnitude of the electric field is
\[E=\sqrt{E_\parallel^2 + E_\perp^2}=4.06~\text{V}/\text{m}\] while its direction is shown in figure (Note that the electron velocity is in the opposite direction with respect to the current.)

A2 ^?? Calculate the difference of the electric potential between the opposite points on the surface of the bar in the direction of the edge $b$.

The potential difference is
\[U_H = E_\perp b = 25~\text{mV}\]

A3 ^?? Find the analytical expression for the DC component of the electric potential difference if the current and the magnetic field are alternating (AC); $I=I_0 \sin \omega t$ and $B= B_0 \sin (\omega t + \delta)$

The potential difference $U_H$ is now time dependent:
\[U_H = \frac{IBb}{ne_0 bc} = \frac{I_0 B_0}{ne_0 c} \sin \omega t \cdot \sin (\omega t + \delta) \] The DC component of $U_H$ is
\[\bar{U}_H = \frac{I_0 B_0}{2 n e_0 c} \cos \delta\]

A4 ^?? Design and explain the electric circuit wich would make possible, be exploting the result A3, to measure the power consumption of an electric apparatus connected with the AC network.

A possible experimental setup

Hall Effect in InSb

Решение