Question

The speed at which electric current travels in a conductor is nearly
$\begin{aligned} & A)3\times {{10}^{4}}m{{s}^{-2}} \\\ & B)3\times {{10}^{5}}m{{s}^{-2}} \\\ & C)3\times {{10}^{6}}m{{s}^{-2}} \\\ & D)3\times {{10}^{8}}m{{s}^{-2}} \\\ \end{aligned}$

Answer 1

Hint : In a conductor, the movement of free electrons in a specific region is known as electric current. These electrons are drifted by an electric field. So if we find the speed of an electric field it will be the same as the speed of current in a conductor. Light is an electro-magnetic wave and we know the speed of light. So the electric field will have the same velocity.

Complete Solution:
When we look at electric current, it is the flow of free electrons through a conductor in an electrical field. This energy travels as an electromagnetic wave and we know the speed of an electromagnetic wave is about$3\times {{10}^{8}}m/s$. So, the speed of current will be the same.
But, there is a small difference because the electrons within the wave move slower than expected. The actual velocity of current is given by the concept of drift velocity. This concept takes account of the collision between these free electrons which results in the slowing of electric current. The drift velocity is directly proportional to the driving electric field and to the average collision time.
But here we will neglect the delay caused by the collision of free electrons inside a conductor.so we will take the speed of current in a conductor is nearly equal to$3\times {{10}^{8}}m/s$.

Therefore, option D is correct.

Note: Here, We have approximated the value of velocity of current in a conductor to be equal to the velocity of the driving field which is equal to speed of light. But actual velocity is somewhat slower due to the inter-electronic collision. The actual velocity is calculated by drift velocity which is given by the expression${{v}_{d}}=\dfrac{1}{2}\left( \dfrac{eE}{m} \right)\tau$.