Hall Effect Magnetic Field Calculation

This problem involves calculating the transverse magnetic field applied to a semiconductor using the Hall effect principle.

Hall Effect Formula

The Hall voltage ($V_H$) generated across a conductor or semiconductor is given by:

$ V_H = \frac{R_H I B}{d} $

Where:

• $V_H$ is the Hall voltage
• $R_H$ is the Hall coefficient
• $I$ is the current flowing through the material
• $B$ is the magnetic field strength perpendicular to the current
• $d$ is the thickness of the material

To find the magnetic field ($B$), we rearrange the formula:

$ B = \frac{V_H d}{R_H I} $

Given Values and Conversions

The provided values are:

• Thickness, $d = 0.7\text{ mm} = 0.7 \times 10^{-3}\text{ m}$
• Hall voltage, $V_H = 0.9\text{ mV} = 0.9 \times 10^{-3}\text{ V}$
• Current, $I = 6\text{ mA} = 6 \times 10^{-3}\text{ A}$
• Hall coefficient, $R_H = 4 \times 10^{-4}\text{ m}^3/\text{C}$

Calculating Magnetic Field (B)

Substitute the converted values into the rearranged formula:

$ B = \frac{(0.9 \times 10^{-3}\text{ V}) \times (0.7 \times 10^{-3}\text{ m})}{(4 \times 10^{-4}\text{ m}^3/\text{C}) \times (6 \times 10^{-3}\text{ A})} $

First, calculate the product in the numerator:

$ V_H \times d = (0.9 \times 10^{-3}) \times (0.7 \times 10^{-3}) = 0.63 \times 10^{-6}\text{ V}\cdot\text{m} $

Next, calculate the product in the denominator:

$ R_H \times I = (4 \times 10^{-4}) \times (6 \times 10^{-3}) = 24 \times 10^{-7}\text{ m}^3\cdot\text{A}/\text{C} $

Now, divide the numerator by the denominator:

$ B = \frac{0.63 \times 10^{-6}}{24 \times 10^{-7}} = \frac{0.63}{24 \times 10^{-1}} = \frac{0.63}{2.4} $

$ B \approx 0.2625\text{ T} $

Result

Rounding the calculated magnetic field to two decimal places gives $0.26\text{ T}$. This value falls within the provided correct answer range.