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:
If by 1.5 MJ you mean 1.5E6 Joules then
W = P t = power X time
W / t = P power
P = 1.5E6 J / 600 sec = 2500 J / s
P = I V
a) I = 2500 J/s / (240 J/c) = 10.4 C / sec = 10.4 amps
b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs
c) E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules
Answer:
The value is 
Explanation:
From the question we are told that
The operating temperature is 
The emissivity is 
The power rating is 
Generally the area is mathematically represented as
Where 
is the Stefan Boltzmann constant with value
So
Electrical energy.................
(1,500 meters) x (1 sec/330 meters) =
(1,500 / 330) (meters-sec/meters) =
4.55 seconds
