91Ó°ÊÓ

List the given parameters: \n1. Number of conduction electrons, \( n = 2 \cdot 10^{21} \) electrons per \( \mathrm{m}^{3} \) \n2. Ribbon width \( b = 0.2 \) cm \n3. Ribbon thickness \( d = 0.005 \) cm \n4. Applied magnetic field \( B = 1 \) kilogauss \n5. Applied voltage across the ribbon \( V = 1 \) V

To keep the units consistent, we need to convert \( b \), \( d \) and \( B \) to meters and Tesla respectively: \n1. Convert \( b \) to meters: \( b = 0.2 \) cm \( \times \) \( 10^{-2} \) \( = 0.002 \) m \n2. Convert \( d \) to meters: \( d = 0.005 \) cm \( \times \) \( 10^{-2} \) \( = 0.00005 \) m \n3. Convert \( B \) to Tesla: \( B = 1 \) kilogauss \( \times \) \( 10^{-4} \) \( = 0.1 \) T

The Hall voltage \( V_H \) across the width of the ribbon can be calculated using the equation: \( V_H = \frac{I \times B \times d}{n \times e} \) where \( I \) is the current, \( B \) is the magnetic field strength, \( d \) is the thickness of the ribbon, \( n \) is the number of charge carriers, and \( e \) is the elementary charge. From Ohm's Law, we know \( I = \frac{V}{R} \), where \( R \) is the resistance. However, we don't have the values for \( I \) and \( R \). We do have the resistivity \( p = 0.016 \) Ohm-m, the applied voltage \( V = 1 \) V, and the dimensions of the ribbon. We can calculate \( R \) using the formula: \( R = p \times \frac{l}{A} \) where \( l \) is the length and \( A \) is the cross-sectional area of the ribbon. Considering that \( l = 0.5 \) cm \( = 0.005 \) m and \( A = b \times d = 0.002 \) m \( \times \) \( 0.00005 \) m \( = 10^{-7} \) \( \mathrm{m}^{2} \), we get \( R = 0.016 \) Ohm-m \( \times \frac{0.005 \) m}{\( 10^{-7} \) \( \mathrm{m}^{2}} \) \( = 0.8 \) Ohm. Now, we can calculate \( I = \frac{V}{R} = \frac{1}{0.8} = 1.25 \) A. Substituting \( I \), \( B \), \( d \), \( n \), and \( e = 1.6 \times 10^{-19} \) C in the equation for \( V_H \), we get \( V_H = \frac{1.25 \times 0.1 \times 0.00005}{2 \times 10^{21} \times 1.6 \times 10^{-19}} = 0.0000049 \) V or 4.9 µV.