Part c is asking for the average drift velocity
displacement = 35 cm
time taken = 30 minutes (or 30 x 60 = 1800 sec)

v=s/t

The question explains that the drift velocity is likely to be very small

Reply 4

OOOllie

OP

11

Thank you so much but what about the rest, I was using the actual drift velocity equation

Reply 5

OOOllie

OP

11

Original post by OOOllie

Thank you so much but what about the rest, I was using the actual drift velocity equation

So what about parts D and e :P

Original post by OOOllie

So what about parts D and e :P

D and E are related.

For D, (IMO) it is a matter of whether you know the formula of current density which is

J = nqv_d

For E, use the definition of current density which is the current per unit area of cross section of the wire

J = \dfrac{I}{A}

It would be good that you go through this webpage to understand the concepts introduced in this problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

Reply 8

OOOllie

OP

11

so I've got the number density as 8.49x10^22 cm3 and 0.019 cm/s so then I converted to get into m and I got 8.49x10^16 and 0.00019
Then because of that I did 8.49x10^16x1.6x10^-19x0.00019 but yet still didn't get it right

Original post by OOOllie

so I've got the number density as 8.49x10^22 cm3 and 0.019 cm/s so then I converted to get into m and I got 8.49x10^16 and 0.00019
Then because of that I did 8.49x10^16x1.6x10^-19x0.00019 but yet still didn't get it right