Answer: I = 111.69 pA
Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.
The hall voltage of a semiconductor sensor is given below as
V = I×B/qnd
Where V = hall voltage = 1.5mV =1.5/1000=0.0015V
I = current =?,
n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3
q = magnitude of an electronic charge=1.609×10^-19c
B = strength of magnetic field = 5T
d = thickness of sensor = 0.8mm = 0.0008m
By slotting in the parameters, we have that
0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008
0.0015 = I×5/7.446×10^-8
I = (0.0015 × 7.446×10^-8)/5
I = 111.69*10^(-12)
I = 111.69 pA
<span>According to the formula :
</span><span>a=<span><span>ΔV / </span><span>ΔT
</span></span></span><span>When a body is moving with a uniform velocity, the acceleration is zero. That's it. You should remember, that velocity is not constant whereas speed is constant.</span>
Answer
Given,
refractive index of film, n = 1.6
refractive index of air, n' = 1
angle of incidence, i = 35°
angle of refraction, r = ?
Using Snell's law
n' sin i = n sin r
1 x sin 35° = 1.6 x sin r
r = 21°
Angle of refraction is equal to 21°.
Now,
distance at which refractive angle comes out
d = 2.5 mm
α be the angle with horizontal surface and incident ray.
α = 90°-21° = 69°
t be the thickness of the film.
So,
t = 2.26 mm
Hence, the thickness of the film is equal to 2.26 mm.
