✔ 最佳答案
At steady state, the electric force and magnetic force acting on a charge carrier balance each other.
Hence, qE = Bqv
where q is the charge on each carrier
E is the electric field intensity
B is the magnetic field intensity
v is the charge drift velocity
But E = V/d
where V is the Hall voltage and d is the width of the semi-conductor slab
hence, qV/d = Bqv
V = Bdv
Since current I = nqvA
where n is the no. of charge carriers per unit volume
A is the cross-sectional area of the slab (= d.t, t is the slab thickness)
Thus, v = I/nqA = I/nqdt
The Hall voltage V = Bd.(I/nqdt) = B.I/nqt
i.e. B = Vnqt/I = (2x10^-5) x (10^25) x (1.6x10^-19) x (1.3x10^-3)/0.4 T
B = 0.104 T
The answer is option A