Question

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Consider a rectangular block of copper whose corners at ( 0, 0, 0 ), ( 3.5e-2, 0,0 ) m, ( 3.5e-2, 1.5e-2, 0 ) m, ( 0, 1.5e-2, 0) m, (0, 0, 2.2e-4 ) m, ( 3.5e-2, 0, 2.2e-4) m, ( 3.5e-2, 1.5e-2, 2.2e-4) mand (0, 1.5, 2.2e-4) m. The block is carrying a current of 3.9 A in the direction of the positive x-axis. There is a magnetic field of strength 0.9 T, directed along the positive z-axis. Calculate the Hall potential difference.Copper has gram molecular weight of 63.5 g and a density of 8920 kg/m3. (I point) 10. are located A. 1.179-6V B. 1.402e-6 V C. O 1.955e-6V D. O 1.776e-6V E. 0.295e-6 V Submit Query

Answer 1

If the current is in the x direction and magnetic field is in the z axis, then the Hall voltage is given by
  
where, z is the thickness of the rectangular bar in the z direction and n is the carrier charge density.
So, first we calculate the carrier charge density as
   $n=\frac{N_A Z \rho_m}{M}$
where, N_A is the avogadro number, Z is the valence electrons per atom ( Z = 1, for copper ) , M is the molecular mass, \rho_m is the mass density of the material. So, for copper with given values, we get
    $n=\frac{6.023\times 10^{23}\times 1\times 8920}{63.5\times 10^{-3}}$
   $\Rightarrow n=8.46\times 10^{28}~electrons/m^3$
And so, for given
      $z=2.2\times 10^{-4}~m$ , I_x = 3.9 A, B_z = 0.9 T, we have the Hall voltage

   $V_H=\frac{3.9\times 0.9}{8.46\times 10^{28}\times 1.602\times 10^{-19}\times 2.2\times 10^{-4}}$
$\Rightarrow V_H= 1.179\times 10^{-6}~V$
So, option (A) is correct.