Question

A $2.5$-m-long, $2.0$-mm-diameter aluminum wire has a potential difference of $1.5V$ between its ends. Consider an electron halfway between the center of the wire and the surface that is moving parallel to the wire at the drift speed. What is the ratio of the electric force on the electron to the magnetic force on the electron?

Short answer

The ratio of the electric force on the electron to the magnetic force on the electron is $\frac{F_E}{F_B}=\boxed{4\times {10}^4}.$

Step-by-step solution

Given Information

We have given that $2.5$-m-long, $2.0$-mm-diameter aluminum wire has a potential difference of $1.5V$between its ends.

Therefore, given values are:

$L=2.5m$

$∆V=1.5V$

role="math" localid="1649445477729" $r=1.0mm\to 1\times {10}^{-3}m$

$$\begin{gathered} \left(n_e\right)Aluminium=6.0\times {10}^{28}{m}^{-3} \\ \rho Aluminium=2.8\times {10}^{-3}\Omega ·m \end{gathered}$$

Simply

The electric force $F_E$is exerted on the charge due to the electric field $E$, so it is given by

localid="1649441289466" $F_E=qE=q\frac{∆V}{l}\left(1\right)$

Where $∆V$is the potential difference and $l$is the length of the wire.

Now, putting the values for $∆V$and $l$into equation $\left(1\right)$to get $F_E$in terms of $q$

$F_E=\frac{q∆V}{l}=\frac{q(1.5V)}{(2.5m)}=(0.6q)N$

Ampere's experiment shows that the magnetic field exerts a magnetic force on the moving charge an it depends on the vector of the velocity.

The magnetic force is given by equation $(29.18)$in the form

$F_B=qv{B}_{wire}=qv\frac{{\mu }_{o}I}{2\pi r}\left(2\right)$

Where $r$is the distance of the electron. We missed the value of $v$and the current $I.$Let us find the current. The wire has a diameter $2$mm, so we get its area by

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{2\times {10}^{-6}m}{2}\right)}^2=3.14\times {10}^{-6}{m}^2$

The resistivity $\rho$depends on the geometry of the material and it is related to the resistance $R,$so we can find $R$by

$R=\frac{\rho L}{A}$

$$\begin{gathered} =\frac{(2.8\times {10}^{-8}\Omega ·m)(2.5m)}{3.14\times {10}^{-6}{m}^2} \\ =2.2\Omega \end{gathered}$$

The wire is connected to a battery with emf $1.5V,$so we use Ohm's law to find the current $I$of the wire by

$I=\frac{\epsilon }{R}=\frac{1.5V}{2.2\Omega }=0.68A$

Calculation

The electron is halfway the distance between the two parallel, so the enclosed current will be divided by $4$

$I=\frac{0.68A}{4}=0.17A$

Now, we want to find the velocity $v.$The drift speed depends on the current density $J$and the charge carrier density $n$and it is given by equation $(29.24)$in the form

$v=\frac{J}{ne}=\frac{1}{ne}\left(\frac{E}{\rho }\right)=\frac{1}{ne}\left(\frac{∆V}{\rho l}\right)=\frac{∆V}{l\rho ne}\left(3\right)$

Where the current density is related to the resistivity by $J=E/\rho$and the electric field is substituted by $E=∆V/l.$Now, we plug the values for $∆V,l,\rho ,n$and $e$into equation $\left(3\right)$to get $v$

$v=\frac{∆V}{l\rho ne}$

$$\begin{gathered} =\frac{1.5V}{(2.5m)(2.8\times {10}^{-8}\Omega ·m)(6\times {10}^{28}{m}^{-3})(1.6\times {10}^{-19}C)} \\ =0.022m/s \end{gathered}$$

Now, we plug the values for $I,v$and $r$into equation $\left(2\right)$to get $F_B$in terms of $q$

$F_B=qv\frac{{\mu }_{o}I}{2\pi r}$

$$\begin{gathered} =q(0.022m/s)\left(\frac{{\mu }_o(0.17A)}{2\pi (0.5\times {10}^{-3}m)}\right) \\ =1.5\times {10}^{-5}qN/C \end{gathered}$$

Now, we get the ratio between $F_E$and $F_B$by

localid="1649446971330" $\frac{F_E}{F_B}=\frac{(0.6q)N}{(1.5\times {10}^{-5})q\hspace{0.17em}N}=\boxed{4\times {10}^4}$