Chapter 19, Problem 123P

(a)

To determine

The drift velocity of electrons.

Answer to Problem 123P

Drift velocity is $1.1\text{mm}/\text{s}$.

Explanation of Solution

Concentration of free electrons is $5.85\times {10}^{28}{\text{m}}^{-3}$, thickness of silver strip is $0.050\text{mm}$, width of strip is $20.0\text{mm}$, magnetic field is $0.80\text{T}$, and current is $10.0\text{A}$.

Write the equation for drift velocity.

$v_d=\frac{I}{ne\left(dw\right)}$

Here, $v_d$ is the drift velocity, $I$ is the current, $n$ is the electron density, $e$ is the charge of electron, $d$ is the thickness of strip, and $w$ is the width of strip.

Conclusion:

Substitute $10.0\text{A}$ for $I$, $5.85\times {10}^{28}{\text{m}}^{-3}$ for $n$, $1.6\times {10}^{-19}\text{C}$ for $e$, $0.050\text{mm}$ for $d$, and $20.0\text{mm}$ for $w$ in the above equation to find $v_d$.

$\begin{array}{c}{v}_d=\frac{10.0\text{A}}{\left(5.85\times {10}^{28}{\text{m}}^{-3}\right)\left(1.6\times {10}^{-19}\text{C}\right)\left(\left(0.050\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\left(20.0\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\right)}\\ =\frac{10.0}{7.692\times {10}^3}\text{mm}\\ =0.0013\text{m}\left(\frac{{\text{10}}^{\text{3}}\text{mm}}{\text{1}\text{m}}\right)\\ =1.3\text{mm}\end{array}$

Therefore, the drift velocity is $1.1\text{mm}/\text{s}$.

(b)

To determine

The hall voltage measured by the meter.

Answer to Problem 123P

Hall voltage is $17\text{μV}$.

Explanation of Solution

Concentration of free electrons is $5.85\times {10}^{28}{\text{m}}^{-3}$, thickness of silver strip is $0.050\text{mm}$, width of strip is $20.0\text{mm}$, magnetic field is $0.80\text{T}$, and current is $10.0\text{A}$.

Write the relation between electric force and magnetic force on electron at equilibrium.

$F_E=F_B$ . (I)

Here, $F_E$ is the electric force and $F_B$ is the magnetic force.

Write the equation for $F_E$.

$F_E=e{E}_H$ . (II)

Here, $E_H$ is the electric field.

Write the equation for $F_B$.

$F_B=e{v}_{d}B$ . (III)

Here, $B$ is the magnetic field.

Rewrite equation (I) by substituting equations (II) and (III).

$\begin{array}{c}e{E}_H=e{v}_{d}B\\ E_H=v_{d}B\end{array}$

Write the equation for hall voltage.

$V_H=E_{H}w$

Here, $V_H$ is the hall voltage.

Rewrite the above relation by substituting $v_{d}B$ for $E_H$.

$V_H=\left(v_{d}B\right)w$

Conclusion:

Substitute $1.1\text{mm}/\text{s}$ for $v_d$, $0.80\text{T}$ for $B$, and $20.0\text{mm}$ for $w$ in the above equation to find $V_H$.

$\begin{array}{c}{V}_H=\left(1.1\text{mm}/\text{s}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\left(0.80\text{T}\right)\left(20.0\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\\ =17\times {10}^{-6}\text{V}\left(\frac{{10}^{-3}\text{μV}}{1\text{V}}\right)\\ =17\text{μV}\end{array}$

Therefore, the hall voltage is $17\text{μV}$.

(c)

To determine

The side of voltmeter with higher potential.

Answer to Problem 123P

The left side of voltmeter is at higher potential.

Explanation of Solution

The higher potential side can be identified from the direction of electron flow. Electrons flow from higher potential side to lower potential side.

Here, electrons towards right side of voltmeter. That is, the left side of potentiometer is at higher potential. Therefore, the left side of voltmeter is at higher potential.