Question

a semiconductor crystal of dimensions 12mm long, 5mm wide, and 1mm thick has a magnetic field fluc density of 0.5wb/m² applied from front to back perpendicular to the largest faces. when a current of 20ma flows lengthwise throught the specimen, the voltage measured across its width is found to be 37microvolt. what is the hall coefficient of this semiconductor

Answer 1

The Hall coefficient of the semiconductor is found using the given voltage, current, crystal dimensions, and magnetic field flux density, resulting in a Hall coefficient of 3.7x10^-9 m³/A·s.

Step-by-step explanation:

The student's question pertains to calculating the Hall coefficient of a semiconductor given the dimensions of the crystal, the magnetic field flux density, the current flowing through the specimen, and the measured Hall voltage. To find the Hall coefficient (RH), we use the formula RH = (VH * d) / (I * B), where VH is the Hall voltage, d is the thickness of the crystal, I is the current, and B is the magnetic field flux density.

First, we convert all the given units to the SI unit system, so the dimensions of the semiconductor become 0.012m x 0.005m x 0.001m, the magnetic field flux density stays as 0.5wb/m², the current is converted to 0.02A, and the voltage is converted to 37x10^-6V. Plugging these values into the formula, we get:

RH = (37x10^-6V * 0.001m) / (0.02A * 0.5wb/m²) = 3.7x10^-9 m³/A·s, which is the Hall coefficient of the semiconductor.