00:02
Hi, in the given problem we are having a copper wire whose length is given as 100 meter and its area of cross -section has been given as 0 .20 millimeter square, which can be written as 0 .20 into 10 dash to par minus 6 meter square.
00:32
Resistivity of the copper here has been given as 1 .7 into 10 to the power minus 8 om into meter.
00:51
So in the first part of the problem we have to find the resistance of a solenoid of an inductor coil made up of this copper wire for which we simply use the expression for the resistance in terms of resistivity which is given as r is equal to row into l by a.
01:17
For resistivity this is 1 .7 into 10 dash to bar minus 8 oam into meter for length this is 100 meter and then divided by the area which is 0 .20 into 10 dash to bar minus 6 meter square.
01:40
This meter into meter will be meter square which will be cancelled by this meter square.
01:46
So the resistance of this wire comes out to be 8 .5 om which becomes the answer for the first part of the problem.
01:59
Now in the second part of the problem we have to find the self -inductance of an inductor coil made up of this copper wire.
02:10
Wire such that it is having total turns as 780.
02:21
It is wound on a cylinder whose diameter was given as 4 .0 centimeter.
02:33
So its radius smaller...