Question

The momentum of electrons is a straight wire of length ${\text{L}}\,{\text{ = }}\,1000{\text{ m}}$ carrying a current ${\text{I = 70 A}}$ , will be (in $N - s$ )
A) $0.40\, \times \,{10^{ - 6}}$
B) $0.20\, \times \,{10^{ - 6}}$
C) $0.80\, \times \,{10^{ - 6}}$
D) $0.16 \times \,{10^{ - 6}}$

Answer 1

Hint: Momentum refers to the quantity of motion that an object has. Here in this question we use the basic momentum formula to solve the answer and after that by putting the values in terms related to electrons, we will get our required answer.

Complete step by step answer:
Given
Length of wire${\text{(L)}}\,{\text{ = }}\,1000{\text{ m}}$
Current ${\text{(I) = 70 A}}$
Now, Momentum of any body is given by ${\text{P}}\,{\text{ = Mv}}$
Here , ${\text{M}}$ is the mass of the body and ${\text{v}}$ is the velocity.
Mass of one electron ${\text{m}} = \,9.1 \times {10^{ - 31}}\,{\text{kg}}$
Charge on one electron $e\, = \,1.6 \times {10^{ - 19}}{\text{C}}$
Number of electrons present in given wire $ = \,\dfrac{{\text{q}}}{e}$
Here, ${\text{q}}$ is the charge on the wire.
Mass of given electrons $ = \,\dfrac{{\text{q}}}{e} \times {\text{m}}$
Now ${\text{q}}$ can be written as ${\text{q = It = }}\dfrac{{{\text{IL}}}}{{\text{v}}}$
Therefore, momentum ${\text{P}}\,{\text{ = Mv }} = \,\dfrac{{\text{q}}}{e}{\text{mv = }}\dfrac{{{\text{ILmv}}}}{{{\text{ve}}}}\, = {\text{ }}\dfrac{{{\text{ILm}}}}{{\text{e}}}$
By putting the given values,
$\dfrac{{{\text{ILm}}}}{{\text{e}}}{\text{ = }}\dfrac{{70 \times 1000 \times 9.1 \times {{10}^{ - 31}}}}{{1.6 \times {{10}^{ - 19}}}}\, = \,0.40\, \times \,{10^{ - 6}}{\text{ N - s}}$
Therefore the correct option is (A).

Note:
We have to remember the values of mass of electron and charge present on one electron. One more point to add here is that force is the multiplication of mass and acceleration but momentum is multiplication of mass and velocity of the body. Momentum has the same direction as that of velocity of the body.