Physics Group - II (3110018)

Here, Magnetic field$= \mathrm{B} = 1.5 \mathrm{Wb}/{\mathrm{m}}^2$
          Electric current$= \mathrm{I} = 200 \mathrm{A}$
          Length of copper strip$= \mathcal{l} = 2 \times {10}^2 \mathrm{m}$
          Density of electron $= \mathrm{n} = 8.4 \times {10}^{28} \mathrm{electrons} / {\mathrm{m}}^3$
          Charge of electron Thickness of the material$= \mathrm{e} = 1.6 \times {10}^{-19} \mathrm{coulomb}$
          Area of thick copper strip $= \mathrm{A} = 2 \times {10}^{-5} {\mathrm{m}}^2$

Now, Hall potential ${\mathrm{H}}_{\mathrm{v}} = \frac{\mathrm{I} \mathrm{B} \mathcal{l}}{\mathrm{n} \mathrm{e} \mathrm{A}}$

                           $\Rightarrow {\mathrm{H}}_{\mathrm{v}} = \frac{\left( 200 \right) \times \left( 1.5 \right) \times \left( 2 \times {10}^{-2} \right)}{\left( 8.4 \times {10}^{28} \right) \times \left( 1.6 \times {10}^{-19} \right) \times \left( 2 \times {10}^{-5} \right)}$

                           $\Rightarrow {\mathrm{H}}_{\mathrm{v}} = 22.32 \times {10}^{-6} \mathrm{V}$