Answer:
a)n= 3.125 x
electrons.
b)J= 1.515 x
A/m²
c)
=1.114 x
m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x
m
radius 'r' = d/2 => 1.025 x
m
no. of electrons 'n'= 8.5 x
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x
C
n= Q/e => 5/ 1.6 x 
n= 3.125 x
electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x
)²)
J= 1.515 x
A/m²
c) The typical speed'
' of an electron is given by:
=
=1.515 x
/ 8.5 x
x |-1.6 x
|
=1.114 x
m/s
d) According to these equations,
J= I/A
=
=
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area
True if you look up the question Is velocity speed in a certain direction you would’ve gotten the answer but I’m pretty sure it’s true
Hi there!
I believe it looks like this.
At positions A, G kinetic energy is maximum.
At position D, potential energy is maximum.
Hope this helps! ☺♥