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
Answer:
303.15K
Explanation:
the formula for converting celsius into kelvin is adding 273.15 to the number
To solve this problem, we will get f and then we will use it to calculate the power.
So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m
Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.
