Question

A silver wire of $ 1nm $ diameter has a charge of $ 90\,coulombs $ flowing in $ 1hours\,15\, minutes $. Silver contains:

Answer 1

Hint: According to the question, when diameter of a wire is given and the charge in it is also given in a given particular time. Then from here, we can find the current flowing in it and we can also find the charge density of the given silver wire.

Complete step by step solution:
Silver wire contains $ 5.8 \times {10^{28}} $ free electrons per $ c{m^3} $ .
Given that:
Diameter of the silver wire is $ 1nm $ .
Silver wire has a charge of $ 90\,coulombs $ .
And, Time taken to flow current-
$t = 1\,hours\,15\, minutes $
$=4500\, seconds $
Now, Current flows is:
 $ i = \dfrac{q}{t} $
Here, $ i $ is electric current.
 $ q $ is the charge.
And, $ t $ is the time taken.
 $ \Rightarrow i = \dfrac{{90}}{{4500}} = 0.02\,amperes $
Now, in this situation, silver wire also contains charge density:
 $ j = \dfrac{i}{A} $
Here, $ j $ is the charge density of the silver wire.
And, $ A $ is the area from where current is flowing.
$\Rightarrow j = \dfrac{{0.02}}{{\pi {r^2}}} $
$\Rightarrow j = \dfrac{{0.02}}{{3.14 \times 0.05 \times 0.05}} $
$ \therefore j = 2.55 \times {10^4}amp.{m^{ - 2}} $
Now, during this phenomenon, another term arises here is Drift Velocity.
$ v(d) = \dfrac{j}{{ne}} $
Here, $ v(d) $ is a drift velocity.
$ j $ is the charge density of the wire.
and, $ n $ is the free electrons moved in a wire.
$\Rightarrow v(d) = \dfrac{{2.55 \times 10000}}{{5.28 \times {{10}^{22}} \times 1.6 \times {{10}^{ - 19}}}} $
$\Rightarrow v(d) = 2.69 \times {10^{ - 7}} $

Note:
We can say that drift velocity of the electrons and its current density is directly proportional to each other. Also, when the electric field intensity increases, the drift velocity increases, and the current flowing through the conductor also increases.