Slides in this set
The following slides contain some worked
examples and how to calculate them.
In them the equation will be manipulated to find
a certain "part" of it like electron drift velocity.
This is originally for OCR Physics specification
but is suitable for other specifications.…read more
Q. Find the drift velocity of electrons of wire diameter of 0.32mm, current of
2.4A and where n is 8.5×10 power of -23 M -3
First we write the equation: A= nAVq and see what parts we got. In the
question we have n, diameter of the wire and the charge. Now we can
rearange the equation:
V= I/ nAq but there is one other step- we need to find the area of the wire as
without it its not possible to find the velocity.
Using formula for area : r squared.
Convert to metres which is simply 0.16×10power of -3 m
Now we use area formula :
×(0.16×10 power of -3)squared = 8.04× power of -8 metres squared. Now that
we have the area we can find the drift velocity!…read more
Now that we have the area its possible to find the drift velocity: consult our
V=I/nAq so we substitute what we got:
V= 2.4A (current)/ (8.5×10 power of 28 the n value)×(8.04×10 power of -8 - area
we calculated) ×(1.6×10 power of -19 electronic charge)= 2.2×10 power of -13
So now we have our answer!!!…read more