Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, [tex]V_{Hall} = 20\ mV = 0.02\ V[/tex]
Magnetic Field, B = 0.10 T
Hall emf, [tex]V'_{Hall} = 69\ mV = 0.069\ V[/tex]
Now,
Drift velocity, [tex]v_{d} = \frac{V_{Hall}}{B}[/tex]
[tex]v_{d} = \frac{0.02}{0.10} = 0.2\ m/s[/tex]
Now, the expression for the electric field is given by:
[tex]E_{Hall} = Bv_{d}sin\theta[/tex] (1)
And
[tex]E_{Hall} = V_{Hall}d[/tex]
Thus eqn (1) becomes
[tex]V_{Hall}d = dBv_{d}sin\theta[/tex]
where
d = distance
[tex]B = \frac{V_{Hall}}{v_{d}sin\theta}[/tex] (2)
(a) When [tex]\theta = 90^{\circ}[/tex]
[tex]B = \frac{0.069}{0.2\times sin90} = 0.345\ T[/tex]
(b) When [tex]\theta = 60^{\circ}[/tex]
[tex]B = \frac{0.069}{0.2\times sin60} = 0.398\ T[/tex]
