Answer:
option D
Explanation:
given,
A conductor is carrying current = 2.0 A is 0.5 mm thick
Hall voltage = 4.5 x 10-6 V
uniform magnetic field = 1.2 T
density of the charge = n =?
hall voltage =[tex]V_h =\dfrac{i\ B}{n\ e\ L}[/tex]
[tex]n = \dfrac{i\ B}{V\ e\ L}[/tex]
[tex]n = \dfrac{2 \times 1.2 }{4.5 \times 10^{-6}\times 1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}[/tex]
n = 6.67 × 10²⁷ charges/m
hence the correct answer is option D
