91Ó°ÊÓ

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" $r=1.0mm→1×{10}^{-3}m$

$$\begin{gathered} \left(n_e\right)Aluminium=6.0×{10}^{28}{m}^{-3} \\ \mathrm{ÒÏ}Aluminium=2.8×{10}^{-3}\mathrm{Î\copyright }·m \end{gathered}$$

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

$F_B=qv{B}_{wire}=qv\frac{{\mathrm{μ}}_{\mathrm{ο}}I}{2\mathrm{π}r}\left(2\right)$

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

$A=\mathrm{π}{\left(\frac{d}{2}\right)}^2=\mathrm{π}{\left(\frac{2×{10}^{-6}m}{2}\right)}^2=3.14×{10}^{-6}{m}^2$

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

$R=\frac{\mathrm{ÒÏ}L}{A}$

$$\begin{gathered} =\frac{(2.8×{10}^{-8}\mathrm{Î\copyright }·m)(2.5m)}{3.14×{10}^{-6}{m}^2} \\ =2.2\mathrm{Î\copyright } \end{gathered}$$

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

$I=\frac{\mathrm{Î\mu }}{R}=\frac{1.5V}{2.2\mathrm{Î\copyright }}=0.68A$

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

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

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

$v=\frac{∆V}{l\mathrm{ÒÏ}ne}$

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

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

$F_B=qv\frac{{\mathrm{μ}}_{\mathrm{ο}}I}{2\mathrm{π}r}$

$$\begin{gathered} =q(0.022m/s)\left(\frac{{\mathrm{μ}}_{\mathrm{ο}}(0.17A)}{2\mathrm{π}(0.5×{10}^{-3}m)}\right) \\ =1.5×{10}^{-5}qN/C \end{gathered}$$

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

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