Question

A neutral metal rod of length 60 cm falls toward the Earth. The rod is horizontal and oriented east west. (1) Which end of the rod, east or west, has excess electrons? Explain using physics diagrams. (2) At a moment when the rod’s speed is 4m/s , approximately how many excess electrons are at the negative end of the rod?

Short answer

$\frac{1.224\times {10}^{14}}{R}$

Step-by-step solution

Given Data

$\begin{array}{c}\text{L=60\hspace{0.33em}cm}\\ \text{speed=4\hspace{0.33em}m/s}\\ \text{earthmagneticfieldB=2}\times {\text{10}}^{-5}\text{\hspace{0.33em}T}\end{array}$

Concept

The measurement of excess electrons is the amount of electrons carried within a charge.

Find the excess electrons

Due to the motion of conducting rod through a magnetic field “B” are induced emf is developed in it.

$\xi \text{\hspace{0.33em}=BLv\hspace{0.33em}sin\hspace{0.33em}}\theta$

Where, earth’s magnetic field acts S - N and conductor is held E - W, so $\theta =90°$

$\begin{array}{c}\xi =BLv\\ =\text{2}\times {\text{10}}^{-5}\times 0.6\times 4\\ =4.8\times {10}^{-5}\end{array}$

Due to this emf an induced current generated in the rod is,

$\begin{array}{c}I=\frac{Ne}{t}=\frac{\xi }{R}\\ \text{where,}\\ t=\frac{v}{g}\\ =\frac{4}{9.8}\\ =0.4082\end{array}$

R is resistance

e is charge of electrons $e=1.6\times {10}^{-19}\text{\hspace{0.33em}C}$

Therefore the excess number of electrons across the conducting rod is,

$\begin{array}{c}N=\frac{\xi t}{eR}\\ =\frac{4.8\times {10}^{-5}\times 0.4082}{1.6\times {10}^{-19}\times R}\\ =\frac{1.224\times {10}^{14}}{R}\end{array}$

Hence, the excess number of electrons is$\frac{1.224\times {10}^{14}}{R}$