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Current (I) = 50A

Thickness of cupper (x) = 0.5cm

Width of copper slab(b) = 2cm

Magnetic Field(B) = 1.5T

Free electron concentration (N) = \(8.4 \times 10^{28}\) electrons/m^{3}

Magnitude of Hall Voltage(V_{H}) = ?

We know that,

\(V_{H} = \frac{IB b}{NqA}\)

\(V_{H} = \frac{50 \times 1.5 \times 2 \times 10^{-2}}{8.4 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-4}}\)

\(V_{H} = 1.12 \times 10^{-6} V\)

Hence required hall voltage is \(1.12 \times 10^{-6} V\).

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