The Hall affect is used to measure the carrier density of a thin sheet of electrons.When a current of 80.0 $\mu$A flows through the length of the electron sheet,which is 1.2-mm long,0.27-mm wide and 12-nm thick,a magnetic field perpendicular to the sheet produces a potential difference of 0.53 mV across the width of the sheet.If the carrier density in the electron sheet is 6.74 * 10²⁵ m^-3,what is the magnitude of the magnetic field?

Question

Quizplus · Accepted Answer

From the Hall effect equation, we have:

B = (VH * t) / (I * w * nq)

where VH is the Hall voltage, t is the thickness of the sheet, I is the current, w is the width of the sheet, n is the carrier density, and q is the electron charge.

Plugging in the given values, we get:

B = (0.53 mV * 12 nm) / (80.0 μA * 0.27 mm * 6.74e10 / (1e9 * 1.6e-19 C))

B = 0.857 T

Therefore, the best choice is B) 0.86 T.

The Hall Affect Is Used to Measure the Carrier Density μ\muμ

The Hall Affect Is Used to Measure the Carrier Density $\mu$