# A current of 50A is established in a slab of copper 0.5cm thick and 2 cm wide .The slab is placed in a magnetic field B of 1.5 T. The magnetic field is perpendicular to the plane of the slab and to the current .The free electron concentration in copper is 8.4 × 1028 electrons/m3? . What will be the magnitude of the gall voltage across the width of the slab?

Solution:

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/m3

Magnitude of Hall Voltage(VH) = ?

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 $$1.12 \times 10^{-6} V$$.